Upper bounds on the difference of consecutive zeta zeros

Question

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.

Answer 1

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 $O( 1/\log\log \gamma_n).$

Answer 2

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 $\varepsilon$.

Answer 3

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.

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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
$$|\gamma-T|<\frac{A}{\log\log\log\;T}.$$
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.