There are many results on the spacing of the gaps between nontrivial zeros of the $\zeta$ function, from trivial (average value is $\frac{2\pi}{\log\gamma_n}$) to difficult (bounds on max and min values of the normalized gap). Are any reasonable upper bounds known? I'd like to have something that says, given any $\varepsilon>0,$ there is some N beyond which the gaps $\gamma_{n+1}-\gamma_n$ is at most $\varepsilon.$ This seems a weak request given the asymptotic behavior but I haven't found anything along these lines.

Any ideas?

I asked the question on math.se but did not get an answer.


Littlewood was the first to prove that the gaps between the ordinates of successive zeros of $\zeta(s)$ tend to zero. This is proved, for instance, in Titchmarsh's book on the zeta-function (see Theorem 9.11).

I believe the best known unconditional result states that $$ \gamma_{n+1}-\gamma_n = O( 1/\log\log\log \gamma_n)$$ as $n\to \infty$. Assuming the Riemann Hypothesis, this can be improved to $O( 1/\log\log \gamma_n).$

  • $\begingroup$ Do you know of an effective version of the first? $\endgroup$ – Charles Jan 6 '12 at 13:26
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    $\begingroup$ This is Theorem 9.12 in Titchmarsh 'Theory of the Riemann Zeta Function.' The proof uses the Borel-Caratheodory theorem, and can be made effective if you really really want it. Titchmarsh has a series of seven successive constants $A_1, A_2, \ldots, A_6, A$, with the final $A$ being the constant in the big Oh term above. $\endgroup$ – Stopple Jan 11 '12 at 22:07

From the standard zero-counting formula $N(T) = \frac{T}{2\pi} \log(\frac{T}{2 \pi e}) + O(\log{T})$, this shows $N(T + h) - N(T) = \frac{h}{2 \pi} \log(\frac{T}{2 \pi}) + O(\log{T})$, and hence $N(T+h) - N(T) \geq 1$ provided $h$ is large enough compared to the implied constants. This shows what you ask for with an unspecified $\varepsilon$.

  • $\begingroup$ Yes, that was my first thought (though I must have do.e my arithmetic wrong since I found a larger error term). But can it be made effective? $\endgroup$ – Charles Jan 5 '12 at 21:00
  • $\begingroup$ It can be made effective, and surely has been, but I don't know a source offhand. $\endgroup$ – Matt Young Jan 5 '12 at 21:15
  • $\begingroup$ This also gives the desired result for some $\varepsilon$ even without the "some $N$ beyond which ...". Can it be proved for every $\varepsilon > 0$ (with $N\phantom.$ necessarily depending on $\varepsilon$)? $\endgroup$ – Noam D. Elkies Jan 5 '12 at 21:54
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    $\begingroup$ Matt, I am a bit confused. Aren't you only proving bounded gaps? $\endgroup$ – Micah Milinovich Jan 6 '12 at 4:58
  • $\begingroup$ @Micah: yes, the question as stated asked for the existence of some $\varepsilon$ for which the gaps are bounded by $\varepsilon$. I know $\varepsilon$ usually wants to tend to zero though... $\endgroup$ – Matt Young Jan 6 '12 at 18:20

Edward Charles Titchmarsh, The theory of the Riemann zeta-function, IX. The general distribution of the zeros pages 191 to 193. Oxford at the Clarendon press 1951.

9.12. We shall now obtain a more precise result of the same kind. $\dagger$
THEOREM 9.12. For every large positive T, $\zeta(s)$ has a zero $\beta+i \gamma$ satisfying
This was first proved by Littlewood by a detailed study of the conformal representation used in the previous proof. This involves rather complicated calculations with elliptic functions. We shall give here two proofs which avoid these calculations.
In the first, we replace the rectangles by a succession of circles. Let $T$ be a large positive number, and suppose that $\zeta(s)$ has no zero $\beta+i \gamma $ such that $T-\delta \leq \gamma \leq T+\delta$, where $\delta < \frac{1}{2}$. Then the function
$$f(s)=\log\zeta(s),$$ where the logarithm has its principal value for $\sigma > 2$, is regular in the rectangle
$$-2 \leq \sigma \leq 3, \;\;\;\;\;\;\; T - \delta \leq t \leq T + \delta$$
$\dagger$ Littlewood (3); proofs given here by Titchmarsh (13), Kramaschke (1).

Let $c_v, C_v, \textbf{C}_v, \Gamma_v$ be four concentric circles, with centre $2-\frac{1}{4}v\delta+iT,$ and radii $\frac{1}{4}\delta,\frac{1}{2}\delta,\frac{3}{4}\delta,\delta$ respectively. Consider these sets of circles for $v=0,1,...,n,$ where $n=[12/\delta]+1,$ so that $2-\frac{1}{4}n\delta \leq -1,$ i.e. the centre of the last circle, or to the left of, $\sigma = -1$. Let $m_v, M_v,$ and $\textbf{M}_v$ denote the maxima of $|f(s)|$ on $c_v, C_v,$ and $\textbf{C}_v$ respectively.

Let $A_1,A_2,...$ denote the constants (it is convenient to preserve their identity throughout the proof). We have $\textbf{R} \{ f(s) \} < A_1\log T$ on all the circles, and $|f(2+iT)|<A_2$. Hence the Borel-Carathéodory theorem for the circles $\textbf{C}_0$ and $\Gamma_0$ gives
$$\textbf{M}_0< \frac{\delta+\frac{3}{4}\delta}{\delta-\frac{3}{4}\delta}(A_1\log T+A_2)=7(A_1 \log T + A_2),$$ and in particular
$$|f(2-\frac{1}{4}\delta+iT)|<7(A_1 \log T + A_2).$$

Hence, applying the Borel-Carathéodory theorem to $\textbf{C}_1$ and $\Gamma_1,$

$$\textbf{M}_1<7\{A_1\log T + |f(2-\frac{1}{4}\delta+iT)|\}<(7-7^2)A_1\log T +7^2 A_2.$$
So generally $$\textbf{M}_v < (7+7^{v+1})A_1 \log T + 7^{v+1}A_2$$ or, say, $$\textbf{M}_v < 7^{v}A_3 \log T. \;\;\;\;\;\;\;\;\;\;\;\;\;(9.12.1)$$ Now by Hadamard's three-circles theorem $$M_v \leq m_{v}^{a}\textbf{M}_{v}^{b},$$

where $a$ and $b$ are positive constants such that $a+b=1;$ in fact $a= \log \frac{3}{2}/\log 3,$ $b=\log 2 / \log 3.$ Also, since the circle $C_{v-1}$ includes the circle $c_v,$ $m_v \leq M_{v-1}.$ Hence

$$M_v \leq M_{v-1}^{a} \textbf{M}_{v}^{b} \;\;\;\;\;\;\;\;(v=1, 2,...,n).$$

Thus $$M_1 \leq M_{0}^{a}\textbf{M}_{1}^{b}, \;\;\;\;\;\;\; M_2 \leq M_{1}^{a}\textbf{M}_{2}^{b} \leq M_{0}^{a^2}\textbf{M}_{1}^{ab}\textbf{M}_{2}^{b},$$ and so on, giving finally $$M_n \leq M_{0}^{a^n}\textbf{M}_{1}^{a^{n-1}b}\textbf{M}_{2}^{a^{n-2}b}...\textbf{M}_{n}^{b}.$$

Hence, by $(9.12.1),$

$$M_n \leq M_{0}^{a^n}7^{a^{n-1}b+2a^{n-2}b+...+nb}(A_3 \log T)^{a^{n-1}b+a^{n-2}b+...+b}.$$

Now $$a^{n-1}+2a^{n-2}b+...+nb<n^2$$ $$a^{n-1}b + a^{n-2}b + ... + b = b(1-a^n)/(1-a)=1-a^n.$$ Hence $$M_n \leq M_{0}^{a^n}7^{n^2}(A_3 \log T)^{1-a^n} < A_4 7^{n^2}(\log T)^{1-a^n},$$ since $M_0$ is bounded as $T \rightarrow \infty.$
But $|\zeta(s)|>t^{A_5}$ for $\sigma \leq -1, \;\; t>t_0,$ so that $M_n>A_5 \log T.$ Hence $$A_5<A_47^{n^2}(\log T)^{-a^n},$$ $$\log \log T < \left(\frac{1}{a}\right)^{n} \left(n^2 \log 7 -\log \frac{A_5}{A_4}\right),$$ $$\log \log \log T < n \log \frac{1}{a} + A_6 \log n,$$ so that $$\delta < \frac{12}{n-1} < \frac{A}{\log \log \log T},$$ and the result follows.

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    $\begingroup$ Wasn't this already given in Micah's answer? $\endgroup$ – Lucia Dec 13 '15 at 16:49

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