---
abstract: 'We show that if the derivative of the Riemann zeta function has sufficiently many zeros close to the critical line, then the zeta function has many closely spaced zeros. This gives a condition on the zeros of the derivative of the zeta function which implies a lower bound of the class numbers of imaginary quadratic fields.'
address: ' [ American Institute of Mathematicsfarmer@aimath.orgDepartment of MathematicsYonsei Universityhaseo@yonsei.ac.kr]{} '
author:
- 'David W. Farmer and Haseo Ki'
title: 'Landau-Siegel zeros and zeros of the derivative of the Riemann zeta function'
---

[^1]

Introduction
============

The spacing between zeros of the Riemann zeta-function and the location of zeros of the derivative of the zeta-function are closely related problems which have connections to other topics in number theory.

For example, if the zeta-function had a large number of pairs of zeros that were separated by less than half their average spacing, one would obtain an effective lower bound on the class numbers of imaginary quadratic fields [@M; @CI]. Also, Speiser proved that the Riemann hypothesis is equivalent to the assertion that the nontrivial zeros of the derivative of the zeta-function, $\zeta'$, are to the right of the critical line [@Sp]. There is a quantitative version of Speiser’s theorem [@LM] which is the basis for Levinson’s method [@L]. In Levinson’s method there is a loss caused by the zeros of $\zeta'$ which are close to the critical line, so it would be helpful to understand the horizontal distribution of zeros of $\zeta'$. The intuition is that the spacing of zeros of the zeta-function should determine the horizontal distribution of zeros of the derivative. Specifically, a pair of closely spaced zeros of $\zeta(s)$ gives rise to a zero of $\zeta'(s)$ close to the critical line. Our main result is a partial converse, showing that *sufficiently many* zeros of $\zeta'(s)$ close to the $\tfrac12$-line implies the existence of many closely spaced zeros of $\zeta(s)$. See Theorem \[t:la\].

We assume the Riemann hypothesis and write the zeros of $\zeta$ as $\rho_j=\tfrac12+i\gamma_j$ and the zeros of $\zeta'$ as $\beta_j'+i\gamma_j'$, where in both cases we list the zeros by increasing imaginary part. We consider the normalized gaps between zeros of $\zeta$ and the normalized distance of $\rho_j'$ to the right of the critical line, given by $$\begin{aligned}
\label{def:lambdas}
\lambda_j=\mathstrut &(\gamma_{j+1}-\gamma_j)\log\gamma_j \cr
\lambda_j'=\mathstrut & (\beta_j'-\tfrac12) \log\gamma_j'.\end{aligned}$$ We are interested in how small the normalized gaps can be, and how small the normalized distance to the critical line can be, so we set $$\begin{aligned}
\lambda=\mathstrut &\liminf_{j\to\infty} \lambda_j\\
\lambda'=\mathstrut &\liminf_{j\to\infty}\lambda_j' .\end{aligned}$$ We also consider the cumulative densities of $\lambda_j$ and $\lambda_j'$, given by $$\begin{aligned}
m(\nu) =\mathstrut & \liminf_{J\to\infty}
\frac{1}{J} \, \#\{j\le J\ :\ \lambda_j \le \nu\}\cr
m'(\nu) =\mathstrut & \liminf_{J\to\infty} \frac{1}{J}\,
\#\{j\le J\ :\ \lambda_j' \le \nu\}.\end{aligned}$$

Soundararajan’s [@S] Conjecture B states that $\lambda=0$ if and only if $\lambda'=0$. This amounts to conjecturing that zeros of $\zeta'(s)$ close to the $\tfrac12$-line can only arise from a pair of closely spaced zeros of $\zeta(s)$. Zhang [@Z] showed that (on RH) $\lambda=0$ implies $\lambda'=0$. Thus, Soundararajan’s conjecture is almost certainly true because $\lambda=0$ follows from standard conjectures on the zeros of the zeta-function, based on random matrix theory.

However, the second author[@K] showed that $\lambda=0$ and $\lambda'=0$ are not logically equivalent. Specifically, Ki[@K] proved

\[thm:ki\] (Haseo Ki [@K]) Assuming RH, $\lambda' >0$ is equivalent to $$\label{eqn:zetacondition}
M(\gamma_j):= \sum_{0<|\gamma_j-\gamma_n|<1} \frac{1}{\gamma_j-\gamma_n}
=O(\log \gamma_j) .$$

Note that the theorem implies Zhang’s result (that $\lambda=0$ implies $\lambda'=0$), because if $\lambda=0$ then for some $j$ the sum in will be large because an individual term in the sum is large. But that is not the only way for $M(\gamma_j)$ to be large. It is possible that there could be an imbalance in the distribution of zeros, such as a very large gap between neighboring zeros, which makes the sum large because many small terms have the same sign.

For example, suppose there were consecutive zeros of the zeta function with a gap of size 1, followed by $c \log T$ zeros equally spaced (this cannot happen, but we are illustrating a point). Then $M(\gamma)$ would be $\gg \log T \log\log T$. That possibility is the reason attempts to prove $\lambda'=0$ implies $\lambda=0$ have been unsuccessful. For example, Garaev and Y[i]{}ld[i]{}r[i]{}m [@GY] required the stronger assumption $\lambda_J'(\log\log \gamma_J')^2=o(1)$ in order to conclude $\lambda_J=o(1)$.

The discussion in the previous paragraph shows that, without detailed knowledge of the distribution of zero spacings, one requires $ M(\gamma)\ge C \log T \log\log T$ for any $C>0$ in order to conclude $\lambda =0$. It is possible that this could be improved by proving results about the rigidity of the spacing between zeros of the zeta function. Random matrix theory could give a clue about the limits of this approach. This would involve finding the expected maximum of the random matrix analogue of the sum $$\label{eqn:zetairregular}
\sum_{\frac{1}{\log \gamma_j}<|\gamma_j-\gamma_n|<1}
\frac{1}{\gamma_j-\gamma_n}.$$ Unfortunately, the necessary random matrix calculation may be quite difficult because a lower bound on $|\gamma_j-\gamma_n|$ requires the exclusion of a varying number of intervening zeros, so the combinatorics of the random matrix calculation may be intricate.

In this paper we consider not $\lambda$ and $\lambda'$, but the density functions $m(\nu)$ and $m'(\nu)$. In the next section we illustrate this with the example described above, and then we state our main result.

Examples with equally spaced zeros {#sec:pictures}
----------------------------------

We illustrate Theorem \[thm:ki\] with examples which can help build intuition for why $\lambda'=0$ does not imply $\lambda=0$.

Our example involves degree $N$ polynomials with all zeros on the unit circle. In other words, characteristic polynomials of matrices in the unitary group $U(N)$. In these examples. $\lambda>0$ but $\lambda'=0$, where $\lambda$ and $\lambda'$ refer respectively to the large $N$ limits of the normalized gap between zeros, and the rescaled distance between zeros of the derivative and the unit circle. This is the random matrix analogue of $\lambda$ and $\lambda'$ for the zeta function.

Figure \[fig:1a\] illustrates the case of 16 zeros in the interval$\{e^{i\theta}\ :\ 0\le \theta\le \pi/2\}$. The plot on the left shows the zeros of the polynomial and its derivative. The figure on the right is the same plot “unrolled”: the horizontal axis is the argument, and the vertical axis is the distance from the unit circle, rescaled by a constant factor.

\[0.7\][![On the left, the zeros and the zeros of the derivative of a degree 16 polynomial having all zeros in $\frac14$ of the unit circle. On the right, the image of those zeros under the mapping $r e^{i \theta} \mapsto (\theta, 2 \pi \cdot 16(1-r))$. Zeros of the function are shown as small squares, and zeros of the derivative as small dots. []{data-label="fig:1a"}](plot1a.eps "fig:")]{} 0.5in \[0.7\][![On the left, the zeros and the zeros of the derivative of a degree 16 polynomial having all zeros in $\frac14$ of the unit circle. On the right, the image of those zeros under the mapping $r e^{i \theta} \mapsto (\theta, 2 \pi \cdot 16(1-r))$. Zeros of the function are shown as small squares, and zeros of the derivative as small dots. []{data-label="fig:1a"}](plot1aT.eps "fig:")]{}

Figure \[fig:1b1c\] is the analogue of the plot on the right side of Figure \[fig:1a\], for 101 zeros and 501 zeros. Note that in these examples $\lambda\sim \pi/2$.

\[0.7\][![Unrolled and rescaled zeros of the derivative of a polynomial with zeros equally spaced along the arc $\{e^{i\theta}\ :\ 0\le \theta\le \pi/2\}$. The polynomial has degree 101 (left) and 501 (right). []{data-label="fig:1b1c"}](plot1bT.eps "fig:")]{} 0.5in \[0.7\][![Unrolled and rescaled zeros of the derivative of a polynomial with zeros equally spaced along the arc $\{e^{i\theta}\ :\ 0\le \theta\le \pi/2\}$. The polynomial has degree 101 (left) and 501 (right). []{data-label="fig:1b1c"}](plot1cT.eps "fig:")]{}

In Figure \[fig:1b1c\] the vertical scales are stretched by a factor of $2 \pi N (1-r)$ where $N=101$ and $501$, respectively.

Figures \[fig:1a\] and \[fig:1b1c\] illustrate that, with this unrolling and rescaling, the zeros of the derivative approach a circle. We see that even though $\lambda>0$ we have $\lambda'=0$, but furthermore, since the zeros lie on a (rescaled) circle, we have $m'(\nu)\gg \nu^2$ as $\nu\to 0$. Thus, we can have $m'(\nu)>0$ for all $\nu>0$, yet $m(\nu)=0$ for $\nu$ sufficiently small.

We believe that the above example is the limit of this behavior, and we make the following conjecture, which we view as a refinement of Soundararajan’s conjecture.

\[conj:conjecture1\] If $m'(\nu) \gg \nu^\alpha$ for some $\alpha<2$, then $m(\nu)>0$ for all $\nu>0$.

We intend this as a general conjecture, applying to the Riemann zeta function but also to other cases such as a sequence of polynomials with all zeros on the unit circle.

For applications to lower bounds of class numbers [@M; @CI] one does not actually need $m(\nu)>0$ for $\nu < \pi $; it is sufficient to show that a relatively small number of gaps between zeros of the zeta function are small. Our main result, Theorem \[t:la\], obtains such bounds from estimates on the zeros of the derivative of the zeta function.

Denote $ \log_{(2)}t=\log\log t $.

\[t:la\] Assume RH. Suppose that for all $\nu>0$, $$\#\{0<\gamma'<T:\left(\beta'-\tfrac{1}{2}\right)\log\gamma'\le\nu\}\geqslant
e^{-C(\nu)}T\log T\qquad(\nu>0,\,T\to\infty),$$ as $T\to \infty$, where $C(\nu)>0$ for $\nu>0$ with $\lim_{\nu\to0^+}\sqrt{\nu}C(\nu)=0$ and $\lim_{\nu\to0^+}C(\nu)=\infty$. Then $$\label{eqn:mainresult}
\liminf_{T\to\infty}\frac{\#\{\gamma_n\le
T:(\gamma_{n+1}-\gamma_n)\log\gamma_n\le\nu\}}{T\log T/\logg
T}>0
$$ for all $\nu>0$.

The conclusion of the theorem is weaker than $m(\nu)>0$ for $\nu>0$, but only by a factor of $\logg T$. Thus, it is more than sufficient to apply the results of Conrey and Iwaniec [@CI]. In particular, Theorem \[t:la\] shows that it is possible to obtain lower bounds for class numbers of imaginary quadratic fields from knowledge of the density of zeros of the derivative of the Riemann zeta function.

There is an apparent discrepancy between Conjecture \[conj:conjecture1\] and Theorem \[t:la\] which we wish to clarify. In Theorem \[t:la\] we allow exponential decrease of $m'(\nu)$ as $\nu\to 0$. While the conclusion of the theorem is weaker than $m(\nu)>0$ by a factor of $\logg T$, it may seem curious that the condition in Conjecture \[conj:conjecture1\] requires $m'(\nu)$ to be relatively large as $\nu\to 0$. Indeed, the examples in Section \[sec:pictures\] show that the condition in Conjecture \[conj:conjecture1\] cannot be improved for general functions.

The reason for the apparent inconsistency is that, as described in Section \[sec:boundMgammac\], our method relies on a bound on the moments of the logarithmic derivative. For the Riemann zeta function one expects $$\label{eqn:zpzbound}
\int_{T}^{2T} \left|
  \frac{\zeta'}{\zeta}\left(\frac12 + \frac{1}{\log T} + i t\right)
\right|^{2k} dt \ll_k T\log^{2k}T.$$ The bound should follow by the method of Selberg [@Sel], although we give a conditional proof that allows us to explicitly determine the implied constant. Such a bound, for one fixed $k$, would establish a weaker version of Theorem \[t:la\] that required $m'(\nu)\gg \nu^{-2 k}$. However, more general functions like the polynomials in Section \[sec:pictures\] do not satisfy an analogous bound to . In fact, they are very large on the unit circle and do not satisfy the analogue of the Lindelöf hypothesis. Conjecture \[conj:conjecture1\] is intended to cover those more general cases, while stronger statements should be true for the zeta function.

It is interesting to speculate on the precise nature of the function $m'(\nu)$ for the Riemann zeta function. Dueñez [*et. al.*]{} [@Due] give a detailed analysis of the relationship between small gaps between zeros of the zeta function (and analogously for zeros of the characteristic polynomial of a random unitary matrix) and the zeros of the derivative which arise from the small gaps. For the case of the Riemann zeta function they indicate that the random matrix conjectures for the zeros of the zeta function should imply $$m_\zeta'(\nu) \sim \frac{8}{9\pi} \nu^{\frac{3}{2}},$$ as conjectured by Mezzadri [@Mez03]. That calculation is based on a more general result which suggests that if $m(\nu)\sim \kappa \nu^\beta$ then $m'(\nu) \sim \kappa' \nu^{\beta/2}$ where $$\kappa'=2\pi \frac{\kappa}{\beta} \left(\frac{2}{\pi}\right)^\beta.$$ The factor of $2\pi$ comes from a different normalization used in [@Due] and here we work with the cumulative distribution functions $m$ and $m'$, while in [@Due] they use density functions. That derivation assumed that zeros of $\zeta'$ close to the $\tfrac12$-line only arise from closely spaced zeros of the zeta-function. The discussion above shows that, without further knowledge of the zeros, this is not a valid assumption. But, as indicated in our Conjecture \[conj:conjecture1\], if $\beta<4$ then we believe that the almost all zeros close to the $\tfrac12$-line do arise in such a manner. The random matrix prediction for the neighbor spacing of zeros of the zeta-function has $\kappa=\pi/6$ and $\beta=3$, which is covered by Conjecture \[conj:conjecture1\]. So our results support the analysis of Dueñez [*et. al.*]{} [@Due].

The remainder of this paper is devoted to the proof of Theorem \[t:la\].

Proof of Theorem \[t:la\]
=========================

Theorem \[t:la\] says that sufficiently many zeros of $\zeta'$ close to the $\frac12$-line can only arise from closely spaced zeros of the zeta-function. If $\rho'=\beta'+i\gamma'$ is a zero of $\zeta'$, then we denote by $\rho_c=\frac12+i\gamma_c$ the zero of the zeta-function which is closest to $\rho'$. Thus, we must show that if there are many $\beta'$ very close to $\tfrac12$, then often there is another zero of the zeta-function close to $\gamma_c$.

Our approach involves a study of the quantity $$M_{\gamma_c}=\sum_{0<|\gamma-\gamma_c|\le
X(\gamma_c)} \frac{1}{\gamma-\gamma'},$$ where the range in the sum, $X(\gamma_c)$, turns out to be a limiting factor in our method. By analogy to a similar quantity studied in [@K], we expect that $M_{\gamma_c}$ should be large if and only if $\beta'-\frac12$ is small. And just like in [@K], there are two ways that $M_{\gamma_c}$ can be large. There could be an individual term which is large. That would happen if $\gamma'$ was near two $\gamma$s that are very close together. Or there could be a large imbalance in the the distribution of the $\gamma$s, for example if there was an unusually large gap between $\gamma_c$ and one of the adjacent zeros. We must show that the second possibility cannot occur too often. This is accomplished by showing that an imbalance in the distribution of zeros causes the zeta function to be large, and bounds on moments of the zeta function show that this cannot happen too often.

The proof involves two steps. Assume the zeros of the zeta function rarely get close together. First we show that if $\beta'-\frac12$ is small then $M_{\gamma_c}$ is large. Second, we show that if $M_{\gamma_c}$ is large then usually $\frac{\zeta'}{\zeta}(s)$ is large near $\tfrac12+i\gamma'$, subject to our assumption that the zeros of the zeta function rarely get close together. Standard bounds for the moments of $\frac{\zeta'}{\zeta}(\sigma+i t)$ let us conclude that $\beta'-\frac12$ cannot be small too often, which is what we wanted to prove.

The relationship between $M_{\gamma_c}$ and $\zeta'/\zeta$ relies on an estimate for $\zeta'/\zeta$ in terms of a short sum over zeros. Suppose we have $$\label{eqn:logderiv1}
\frac{\zeta'}{\zeta}(s) = \sum_{|\gamma-t|< X(T)} \frac{1}{s-\rho} + O(\log T).$$ On RH, with $X(t)=1/\logg T$ the above holds for all $t$ [@Ti]. Using this, instead of our  below, leads to a weaker version of Theorem \[t:la\], where the $\logg T$ in the denominator of is replaced by $\log T$.

We prove the following strengthening of , but only near almost all $\gamma$.

\[p:logdea\] Assume RH. Let $m_0$ be a positive integer. If $C^*>1$ is sufficiently large, then the number of $\gamma_n<T$ such that $$\label{e:logdea}
\frac{\zeta'}{\zeta}(s)=
\sum_{|\gamma-t|\le
\frac{C^*\logg\gamma}{\log\gamma}}\frac{1}{s-\rho}
+
O(\log\gamma_n)$$ for $s=1/2+1/\log\gamma_n+it$ with $t\geqslant10$ and $|\gamma_n-t|\le A/\log\gamma_n$ is $$\frac{T}{2\pi}\log T+O\left(\frac{T}{(\log T)^{m_0}}\right)$$ as $T\to\infty$.

The proof of Proposition \[p:logdea\] is in Section \[sec:logdea\].

Restricting to zeros with special properties
--------------------------------------------

We begin the proof of Theorem \[t:la\]. The lemmas in this section show that, in the context of the proof of Theorem \[t:la\], we only have to deal with zeros that are well spaced.

Suppose, for the purposes of contradiction, that there exists $\epsilon>0$ so that $$\label{eqn:nosmallgaps}
\liminf_{T\to\infty}\frac{\#\{\gamma_n\le
T:\gamma_{n+1}-\gamma_n\le\epsilon/\log\gamma_n\}}{T\log
T/\logg T}=0.$$ Then, we can find a sequence $\langle T_l\rangle$ such that $T_1$ is sufficiently large, $T_l\to\infty$ and $$\#\{\gamma_n\le
T_l:\gamma_{n+1}-\gamma_n\le\epsilon/\log\gamma_n\}=o\left(T_l\log
T_l/\logg T_l\right)$$ as $l\to\infty$. We set $$T=T_l.  $$

The following lemma shows that we can restrict our attention to those zeros whose immediate neighbors are well spaced.

\[lem:density\] Let $K=4C^*\left[\logg T\right]$. Under assumption  we have $$\label{e:density}
\# \{\gamma_n<T:0<|m|\le
K,\,|\gamma_{n+m}-\gamma_{n+m-1}|\geqslant\frac{\epsilon}{2\log\gamma_n}\}=\frac{T}{2\pi}\log
T(1+o(1)).$$

For each $m=\pm1,\pm2,\ldots$, let $$A_m=\{\gamma_n<T:|\gamma_{n+m}-\gamma_{n+m-1}|\geqslant\frac{\epsilon}{2\log\gamma_n}\}.$$ Here, we exclude the case $n+m\le1$. By assumption  have $$\#(A_m)=\frac{T}{2\pi}\log T+o\left(\frac{T\log T}{\logg T}\right)$$ for $0<|m|\le\log T$. We see that $$\aligned \#\left(\bigcap_{0<|m|\le
K}A_m\right)=&\sum_{0<|m|\le
K}\#(A_m)-\sum_{\substack{-K\le
m<K\\ m\not=0}}\#\left(A_m\cup\bigcap_{\substack{m<l\le K\\
l\not=0}}A_l\right)\\
\geqslant&2K\frac{T}{2\pi}\log T+o\left(\frac{KT\log T}{\logg
T}\right)-(2K-1)\frac{T}{2\pi}\log
T+O(K\log T)\\
=&\frac{T}{2\pi}\log T+o(T\log T).
\endaligned$$

The next Proposition shows that we can restrict to intervals where the number of zeros is close to its average. Fix $C^*>1$, let $l_1$ and $l_2$ be integers, and for $\tfrac12+i\gamma$ a zero of the zeta function set $$N(\gamma,l_1,l_2)=
N\left(\gamma+\frac{l_2C^*\logg\gamma}{\log\gamma}\right)
-N\left(\gamma+\frac{l_1C^*\logg\gamma}{\log\gamma}\right)
-\frac{(l_2-l_1)C^*\logg\gamma}{2\pi}$$ Using an argument in [@Iv], we get the following.

\[p:dnumber\] Let $m_0>0$. There exists $C>0$ such that the number of $\gamma_n<T$ with $$N(\gamma_n,l_1,l_2)\le C\logg T$$ is $$\frac{T}{2\pi}\log T+O\left(\frac{T}{(\log T)^{m_0}}\right)\qquad(T\to\infty),$$ provided that $|l_1|,|l_2|\le\log T/(C^*\logg T)$ and $0<l_2-l_1\le2\log T/(C^*\logg T)$.

The proof of Proposition \[p:dnumber\] is in Section \[sec:dnumber\].

Lower bound for $M_{\gamma_c}$
------------------------------

Let $\beta'+i\gamma'$ be a zero of $\zeta'$, and (assuming RH) let $\tfrac12+i\gamma_c$ be the zero of the zeta function which is closest to $i\gamma'$. If there are two closest zeros, choose the one nearer to the origin. We will use the above lemmas to give a lower bound for $M_{\gamma_c}$, assuming $\beta'-\tfrac12$ is small.

Let $Z(T)$ be the set of $\gamma_c < T$ which satisfy the following three conditions: $$\begin{aligned}
\gamma_c\in\mathstrut & \{\gamma_n<T :
    0<|m|\le K,\,
    |\gamma_{n+m}-\gamma_{n+m-1}|\geqslant
            \frac{\epsilon}{2\log\gamma_n}\};
\label{eqn:gammacset}\\
N(\gamma_c,l_1,l_2)\le\mathstrut & 
C\logg T\qquad
    \left(-\frac{\log T}{C^*\logg T}\le l_1<l_2\le\frac{\log T}{C^*\logg T}\right);
\label{eqn:gammacN}\\
\frac{\zeta'}{\zeta}(s)=\mathstrut & 
   \sum_{|\gamma-t|\le\frac{\logg\gamma}{\log\gamma}}
    \frac{1}{s-\rho}
\label{eqn:gammaczpz}
+
O(\log \gamma_c),\end{aligned}$$ where $s=1/2+1/\log\gamma_c+it$ and $|\gamma_c-t|\le
A/\log\gamma_c$. By the lemmas in the previous section, as $T\to\infty$ the set $Z(T)$ contains $\sim \frac{1}{2\pi}T\log T$ elements. For the remainder of the proof we will assume $\gamma_c\in Z(T)$.

Recall Titchmarsh [@Ti], Theorem 9.6(A): $$\label{eqn:zpzeasy}
\frac{\zeta'}{\zeta}(s)=-\frac{1}{2}\log
t+O(1)+\sum_{\rho}\left(\frac{1}{s-\rho}-\frac{1}{\rho}\right),$$ uniformly for $t\geqslant10$ and $-1\le\sigma\le2$. Let $\beta'+i\gamma'$ be a zero of $\zeta'(s)$ where $0<\gamma'<T$ is sufficiently large. Taking the real part we have $$\label{e:gprime}
\frac{1}{2}\log\gamma'+O(1)
=
\frac{\beta'-\frac{1}{2}}
    {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2}
+
\sum_{\gamma\not=\gamma_c}
\frac{\beta'-\frac{1}{2}}
    {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma)^2}.$$

There are three cases to consider.

[**Case 1.**]{} $\beta'-1/2>|\gamma'-\gamma_c|$.

Then, by (\[e:gprime\]), we get $$\frac{1}{2}\log\gamma'\geqslant\frac{1}{2\left(\beta'-\frac{1}{2}\right)}.$$ Thus, we have $\beta'-1/2 \gg 1/\log\gamma'$.

[**Case 2.**]{} $\beta'-1/2\le|\gamma'-\gamma_c|$ and $|\gamma'-\gamma_c|>\delta(\epsilon)/\log\gamma'$, where $\delta(\epsilon)=8/\epsilon^2$.

By , , and , we have $$\aligned
\frac{1}{2}\log\gamma'\ll&\left(\beta'-\frac{1}{2}\right)\log^2\gamma'+\sum_{m=1}^{\infty}
\frac{\beta'-\frac{1}{2}}{\left(\frac{m\epsilon}{\log\gamma'}\right)^2}+\sum_{m=0}^{\infty}
\frac{\left(\beta'-\frac{1}{2}\right)\logg\gamma'}
{\left(\frac{\logg\gamma'}{\log\gamma'}\right)^2+\left(\frac{m\logg\gamma'}{\log\gamma'}\right)^2}\\
\ll&\left(\beta'-\frac{1}{2}\right)\log^2\gamma'
\endaligned$$ and so again we have $$\beta'-\frac{1}{2}\gg\frac{1}{\log\gamma'}.$$ Here the implied constants depend only on $\epsilon$.

[**Case 3.**]{} $\beta'-1/2\le|\gamma'-\gamma_c|$ and $|\gamma'-\gamma_c|\le\delta(\epsilon)/\log\gamma'$.

Using , , and , as in Case 2, we get $$\frac{1}{2}\log\gamma'\geqslant\frac{\beta'-\frac{1}{2}}{2(\gamma'-\gamma_c)^2}$$ $$\frac{1}{2}\log\gamma'\ll\frac{\beta'-\frac{1}{2}}{(\gamma'-\gamma_c)^2}+\left(\beta'-\frac{1}{2}\right)\log^2\gamma'\\
\ll\frac{\beta'-\frac{1}{2}}{(\gamma'-\gamma_c)^2}.$$ Thus we have $$\label{eqn:twosided}
(\gamma'-\gamma_c)^2\log\gamma'\ll
\beta'-\frac{1}{2}\ll(\gamma'-\gamma_c)^2\log\gamma'.$$ Here the implied constants depend only on $\epsilon$. By  and the conditions of Case 3 we have $$\label{eqn:nosum}
\frac{\gamma_c-\gamma'}{\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2}
    -\frac{1}{\gamma_c-\gamma'}=O(\log\gamma').$$

Now take the imaginary part of  to get $$\label{eqn:sum1}
\sum_{0<|\gamma-\gamma_c| \le \frac{C^*\logg\gamma_c}{\log\gamma_c}}
\frac{\gamma-\gamma'}{\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma)^2}
+\frac{\gamma_c-\gamma'}
    {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2}
= O\left(\log\gamma'\right).$$

Finally, by we have $$\begin{aligned}
\label{eqn:Mcsum}
\sum_{0<|\gamma-\gamma_c|\le
\frac{C^*\logg\gamma_c}{\log\gamma_c}}
    \frac{\gamma-\gamma'}{\left(\beta'-\frac{1}{2}\right)^2
    +(\gamma'-\gamma)^2}-M_{\gamma_c}
=\mathstrut&
\sum_{k=1}^{\infty}\frac{\left(\beta'-\frac{1}{2}\right)^2}{\left(\frac{\epsilon k}{\log\gamma'}\right)^3}\\
=\mathstrut&O(\log\gamma'),\end{aligned}$$ where $$M_{\gamma_c}=\sum_{0<|\gamma-\gamma_c|\le
\frac{C^*\logg\gamma_c}{\log\gamma_c}}\frac{1}{\gamma-\gamma'}.$$ By combining , , , and , we have $$O(\log\gamma')=M_{\gamma_c}+\frac{1}{\gamma_c-\gamma'}=M_{\gamma_c}
+ A_{\gamma_c} \sqrt{\frac{\log\gamma'}{\beta'-\frac{1}{2}}},$$ where $1\ll A_{\gamma_c}\ll 1 $, with the implied constants depending only on $\epsilon$.

Let $\nu$ be a positive number. Suppose that $$\left(\beta'-\frac{1}{2}\right)\log\gamma'\le\nu.$$ Then, for sufficiently small $\nu$, we see that only Case 3 is possible for sufficiently large $\gamma'$, namely we have $$M_{\gamma_c}+
     A_{\gamma_c} \sqrt{\frac{\log\gamma'}{\beta'-\frac{1}{2}}}
=O(\log \gamma').$$ By this, the assumption in Theorem \[t:la\], and the fact that $\#Z(T)\sim \frac{1}{2\pi} T\log T$, we have $$e^{-C(\nu)}\le\frac{1}{\frac{T}{2\pi}\log
T}\#\{0<\gamma'<T: \gamma_c\in Z(T) \text{ and } |M_{\gamma_c}|\gg \frac{\log\gamma'}{\sqrt{\nu}}(1+O(\sqrt{\nu}))\}.$$ By the last inequality we have $$\label{e:nui}
\frac{e^{-C(\nu)}\log^{2k}T}{\nu^{k}}(1+O(\sqrt{\nu}))^{2k}\frac{T}{2\pi}\log
T\ll\sum_{\substack{\frac{T}{\log T}\le\gamma'\le T\\
    \left(\beta'-\frac{1}{2}\right)\log\gamma'\le\nu\\
    \gamma_c\in Z(T)}}
|M_{\gamma_c}|^{2k}.$$

In the next section we describe upper bounds for the moments of $M_{\gamma_c}$. This will contradict  and complete the proof of Theorem \[t:la\].

Bounding the moments of $M_{\gamma_c}$ {#sec:boundMgammac}
--------------------------------------

We obtain an upper bound on $M_{\gamma_c}$ from a bound on moments of the logarithmic derivative of the zeta function. This makes use of that fact that, assuming the zeros of the zeta function do not get close together, the logarithmic derivative can be approximated either by a short sum over zeros, or by a short Dirichlet series.

\[l:moment\] Assume RH. Let $\gamma'<T$ such that $|\gamma'-\gamma_c|\le\delta(\epsilon)/\log\gamma'$ and assume – . Then $$\frac{M_{\gamma_c}}{i}+\frac{\zeta'}{\zeta}\left(\frac{1}{2}+\frac{1}{\log
T}+it\right)=O_\epsilon (\log T),$$ for $|t-\gamma'|\le A/\log\gamma'$.

By the assumptions we have $$\begin{aligned}
\frac{M_{\gamma_c}}{i}+\frac{\zeta'}{\zeta}\left(\frac{1}{2}+\frac{1}{\log
T}+it\right)
=\mathstrut&\frac{M_{\gamma_c}}{i}+\sum_{0<|\gamma-\gamma_c|\le
\frac{C^*\logg\gamma_c}{\log\gamma_c}}\frac{1}{\frac{1}{\log T}+i(t-\gamma)}
    +O(\log T)\\
=\mathstrut&\sum_{0<|\gamma-\gamma_c|\le
\frac{C^*\logg\gamma_c}{\log\gamma_c}}
    \frac{(\frac{1}{\log T}+i (t-\gamma))}
    {(\gamma-\gamma')(\frac{1}{\log T}+i(t-\gamma))}
+O(\log T)\\
\ll\mathstrut &\sum_{m=1}^{\infty}
     \frac{\frac{1+\delta(\epsilon)}{\log\gamma'}}
        {\left(\frac{m\epsilon}{\log\gamma'}\right)^2}+O(\log T)\\
=\mathstrut &O_\epsilon (\log T).\end{aligned}$$

\[lem:zpzapprox\] Assume RH and – . Let $s=\frac12+\frac{1}{\log T}+i t$ with $|t|\le T$, and let $x=T^{1/100 k}$. Then if $|\gamma'-\gamma_c|\le\delta(\epsilon)/\log\gamma'$ and $|t-\gamma'|\le \epsilon /\log\gamma'$, we have $$\frac{\zeta'}{\zeta}(s)=-\sum_{n<x^2}\frac{\Lambda_x(n)}{n^s}+O_\epsilon(k\log T),$$ where $$\Lambda_x(n)=
\begin{cases}
\Lambda(n) & 1\le n\le x \cr
\Lambda(n)\frac{\log(\frac{x^2}{n})}{\log x} & x\le n\le x^2
\end{cases} .$$

By [@Ti], Theorem 14.20, $$\begin{aligned}
\frac{\zeta'}{\zeta}(s)=\mathstrut &-\sum_{n<x^2}\frac{\Lambda_x(n)}{n^s}+\frac{x^{2(1-s)}-x^{1-s}}{(1-s)^2\log
x}\cr
&+ \frac{1}{\log
x}\sum_{q=1}^{\infty}\frac{x^{-2q-s}-x^{-2(2q+s)}}{(2q+s)^2}+\frac{1}{\log
x}\sum_{\rho}\frac{x^{\rho-s}-x^{2(\rho-s)}}{(s-\rho)^2}.\end{aligned}$$ The assumptions on the zero spacings give the claimed bound on the terms involving the zeros.

\[lem:sound\] (Soundararajan, Lemma 3 of [@So1]) Let $T$ be large, and let $2\le x\le T$. Let $k$ be a natural number such that $x^k\le T/\log T$. For any complex numbers $a(p)$ we have $$\int_T^{2T}\left|\sum_{p\le x}\frac{a(p)}{p^{\frac{1}{2}+it}}\right|^{2k}dt\ll
k!\, T \left(\sum_{p\le x} \frac{|a(p)|^2}{p}\right)^{k} ,$$ where the sum is over the primes.

We assemble the above lemmas to bound the moments of $M_{\gamma_c}$.

By Lemma \[l:moment\] and Lemma \[lem:zpzapprox\], with $A=A_\epsilon$ a constant depending only on $\epsilon$, which may be different in each inequality, we have $$\begin{aligned}
\label{eqn:Mgammacbound}
|M_{\gamma_c}|^{2k} \ll \mathstrut & A^{2k} \log^{2k}T +  2^{2k} \left|
        \frac{\zeta'}{\zeta}\left(\frac12 + \frac{1}{\log T} + i t\right)
   \right|^{2k} \cr
\ll \mathstrut & A^{2k} k^{2k} \log^{2k}T + 
2^{2k}
\left|\sum_{n<x^2}
   \frac{\Lambda_x(n)}{n^{\frac12 + \frac{1}{\log T} + i t}}\right|^{2k}\cr
\ll \mathstrut & A^{2k} k^{2k} \log^{2k}T + 
2^{2k}
\left|\sum_{p<x^2}
   \frac{\Lambda_x(p)}{p^{\frac12 + \frac{1}{\log T} + i t}}\right|^{2k},\end{aligned}$$ where $x=T^{1/100k}$, for $|t-\gamma'|\le \delta(\epsilon)/\log\gamma'$, provided $\gamma_c$ satisfies – . That is, provided $\gamma_c\in Z(T)$.

Integrating inequality  over the set $$\left\{T/{\log T} <t < T\ :\ 
|t-\gamma_c|< \delta(\epsilon)/\log T \text{ for some } \gamma_c\in Z(T) \right\}$$ and then using Lemma \[lem:sound\] we get $$\begin{aligned}
\frac{\delta(\epsilon)}{\log T} \sum_{\substack{\frac{T}{\log T}\le\gamma'\le T\\
    \left(\beta'-\frac{1}{2}\right)\log\gamma'\le\nu \\
    \gamma_c\in Z(T)}}
|M_{\gamma_c}|^{2k}
\ll \mathstrut & A^{2k} k^{2k} T\log^{2k}T +
   2^{2k} \int_{\frac{T}{\log T}}^T
      \left| 
\sum_{p<x^2}
   \frac{\Lambda_x(p)}{p^{\frac12 + \frac{1}{\log T} + i t}}
      \right|^{2k}dt\cr
\ll  \mathstrut& A^{2k} k^{2k} T\log^{2k}T 
+ 2^{2 k} k!\, T \left( \sum_{p<x^2}
   \frac{\Lambda_x(p)^2}{p^{1 + \frac{2}{\log T} }}
      \right)^{k} \cr
\ll  \mathstrut& A^{2k} k^{2k} T\log^{2k}T.\end{aligned}$$ The last step used $\Lambda_x(p)\le \Lambda(p)$ and the fact that $$\sum_{p\le x}\frac{\Lambda(p)^2}{p} \ll \log^2 x,$$ which is a weak form of the prime number theorem.

Rearranging the above inequality and combining with (\[e:nui\]), we have $$\frac{e^{-C(\nu)}}{\nu^k}(1+O(\sqrt{\mathstrut\nu}))^{2k}T\log^{2k+1}T
\ll  A^{2k} k^{2k} T\log^{2k+1}T,$$ which rearranges to give $$\left(1+O(\sqrt{\mathstrut\nu})\right)^{2k}
\ll A^{2k} k^{2k} \nu^k e^{C(\nu)}.$$ Letting $k=[1/\sqrt{A^2 e\nu}]$, we have a contradiction if $\sqrt{\mathstrut\nu}C(\nu) \to 0$ as as $\nu\to 0$. This completes the proof of Theorem \[t:la\].

Proofs of technical results
===========================

In this section we provide the proofs of Proposition \[p:logdea\] and Proposition \[p:dnumber\].

Proof of Proposition \[p:dnumber\] {#sec:dnumber}
----------------------------------

A special case of the Proposition is the following:

There exists $C_1>0$ such that the number of $\gamma_n<T$ satisfying $$N\left(\gamma_n+\frac{lC^*\logg T}{\log
T}\right)-N\left(\gamma_n+\frac{(lC^*-1)\logg T}{\log
T}\right)\le C_1\logg T,$$ for all $|l|\le\log T/(C^*\logg T)$, is $$\frac{T}{2\pi}\log T+O\left(\frac{T}{(\log T)^{m_0}}\right).$$ Here $C_1$ is not depending on $C^*$.

The proof of Claim follows easily from the same method below. Thus, we omit the proof of it.

From now on, we are assuming that $\gamma_n$ satisfies $T/(\log
T)^{m_0+1}<\gamma_n<T$ and Claim. We recall $$\begin{aligned}
\int_T^{T+H}|S(t+h)-S(t)|^{2k}dt=\mathstrut&\frac{H(2k)!}{(2\pi^2)^kk!}\log^k(2+h\log
T)\\
&+O\left(H(ck)^k\left(k+\log^{k-1/2}(2+h\log T)\right)\right)\end{aligned}$$ uniformly for $T^a<H\le T$, $a>1/2$, $0<h<1$ and any positive integer $k$, where $c$ is a positive constant and $S(t)=\frac{1}{\pi}\arg\zeta(1/2+it)$. For this, see [@Ts Theorem 4]. Thus we have $$\label{e:tsang}
\int_0^T|S(t+h)-S(t)|^{2k}dt\ll T\left(Ak\right)^{2k},$$ where $\log(2+h\log T)\ll k$. We note that $$\label{e:ost}
S(t+h)-S(t)=N(t+h)-N(t)-\frac{h}{2\pi}\log
t+O\left(\frac{h^2+1}{t}\right),$$ where $N(t)$ is the number of zeros of $\zeta(s)$ in $0<\Im s<t$. By this, we have $$\widetilde{S}(t,l_1,l_2)=N\left(t+\frac{(l_2-l_1)C^*\logg T}{\log
T}\right)-N(t)-\frac{(l_2-l_1)C^*\logg
T}{2\pi}+O\left(\frac{1}{t}\right),$$ where $$\widetilde{S}(t,l_1,l_2)=S\left(t+\frac{(l_2-l_1)C^*\logg T}{\log
T}\right)-S(t)$$ Using Claim, the last formula and (\[e:ost\]), we have $$\begin{aligned}
N(n,l_1,l_2)\le\mathstrut &\left|\widetilde{S}(t,l_1,l_2)\right|+\logg T\\
\le\mathstrut &\left|\widetilde{S}(t-h,l_1,l_2)\right|+3\logg T\\
&\mathstrut +\sum_{j=1}^{2}N\left(\gamma_n+\frac{l_jC^*\logg T}{\log
T}\right)-N\left(\gamma_n+\frac{(l_jC^*-1)\logg T}{\log T}\right)\\
\le\mathstrut &C_2\logg T+\left|\widetilde{S}(t-h,l_1,l_2)\right|\end{aligned}$$ for $t=\gamma_n+l_1C^*\logg T/\log T$ and $0\le
h\le\logg T/\log T$, where $C_2=\max\{2C_1+3,A\}$. Using this, we have $$\begin{aligned}
\sum_{\substack{\frac{T}{(\log T)^{m_0+1}}<\gamma_n<T\\
N(n,l_1,l_2)\geqslant C\logg T}}(C\logg T)^{2k}
\mathstrut & \mathstrut \frac{\logg T}{\log
T}\\
\ll \mathstrut &T\log T(2C_2\logg
T)^{2k}+\sum_{\gamma_n<T}\int_{\gamma_n+\frac{(l_1C^*-1)\logg
T}{\log T}}^{\gamma_n+\frac{l_1C^*\logg T}{\log
T}}\left|2\widetilde{S}(t,l_1,l_2)\right|^{2k}dt\\
\ll \mathstrut & T\log T(2C_2\logg T)^{2k}
    +\log T\int_0^T\left|2\widetilde{S}(t,l_1,l_2)\right|^{2k}dt\\
\ll \mathstrut &T\log T\left((2C_2\logg T)^{2k}+(2C_2k)^{2k}\right)\end{aligned}$$ for any sufficiently large $T$ and any $|l_1|,|l_2|\le\log
T/(C^*\logg T)$ with $0<l_2-l_1\le 2\log T/(C^*\logg T)$. We put $$k=[\logg T]\qquad\text{and}\qquad C=e^{m_0+2}(2C_2+1).$$ By these and the last inequality, we have $$\sum_{\substack{|l_1|,|l_2|\le\frac{\log T}{C^*\logg
T}\\0<l_2-l_1\le\frac{2\log T}{C^*\logg T}}}
\sum_{\substack{\frac{T}{(\log T)^{m_0+1}}<\gamma_n<T\\
N(n,l_1,l_2)\geqslant C\logg T}}1\ll\frac{T(\log T)^4(2C_2\logg
T)^{2k}}{\left(C\logg T\right)^{2k}}\ll\frac{T}{(\log T)^{m_0}}.$$ We complete the proof of Proposition \[p:dnumber\].

Proof of Proposition \[p:logdea\] {#sec:logdea}
---------------------------------

We recall $$\frac{\zeta'}{\zeta}(s)=O(\log
t)+\sum_{|\gamma-t|\le1}\frac{1}{s-\rho}$$ holds uniformly for $t>1$ and $-2\le{\mathbb R}s\le1$. For this, see [@Ti Theorem 9.6 (A)]. Using the last formula, it suffices to show that the number of $\gamma_n$ in $T/(\log
T)^{m_0+1}\le\gamma_n<T$ such that $\gamma_n$ satisfies the condition in Proposition \[p:dnumber\] and $$\sum_{\frac{C^*\logg T}{\log
T}<|\gamma_n-\gamma_m|\le1}\frac{1}{\gamma_n-\gamma_m}=O(\log
T)$$ is $$\frac{T}{2\pi}\log T+O\left(\frac{T}{(\log
T)^{m_0}}\right)\qquad(T\to\infty),$$ because for $s=1/2+1/\log T+it$ and $|\gamma_n-t|\le A/\log
T$, we have $$\sum_{\frac{C^*\logg T}{\log
T}<|\gamma_n-\gamma_m|\le1}\frac{1}{s-\rho}-\frac{1}{i(\gamma_n-\gamma_m)}=O\left(\frac{1}{\log
T}\sum_{m=1}^{\infty}\frac{\logg T}{\left(\frac{m\logg T}{\log
T}\right)^2}\right)=O(\log T).$$ We recall that Proposition \[p:dnumber\] implies $$N\left(\gamma_n+\frac{lC^*\logg T}{\log T}\right)
=N(\gamma_n)+\frac{lC^*\logg T}{2\pi}+O\left(\logg T\right)$$ for any integer $l$ with $|l|\le\log T/(C^*\logg T)$. This immediately implies that for a sufficiently large $C^*>1$, we have $$\max_{0\le k\le
N}|2\gamma_n-\gamma_{m_2+k}-\gamma_{m_1-k}|\frac{\log
T}{2\pi}=O(\logg T),$$ where $\gamma_{m_1}$ is the greatest one in $[\gamma_n-1,\gamma_n-C^*\logg T/\log T)$, $\gamma_{m_1}$ the least one in $(\gamma_n+C^*\logg T/\log T,\gamma_n+1]$ and $N$ the largest positive integer such that $\gamma_{m_1-N}$ and $\gamma_{m_2+N}$ belong to $[\gamma_n-C^*\logg T/\log T,\gamma_n+C^*\logg T/\log T]$. By this and putting $ M(n)=\max_{0\le k\le
N}|2\gamma_n-\gamma_{m_2+k}-\gamma_{m_1-k}|,$ we have $$M(n)\ll\frac{\logg T}{\log T}$$ Using this and Proposition \[p:dnumber\] and the fact [@Ti Theorems 9.3 and 14.13] that the number of zeros between $t$ and $t+1$ is $$\frac{\log t}{2\pi}+O\left(\frac{\log t}{\logg
t}\right)\qquad(\text{as } t\to\infty),$$ we have $$\begin{aligned}
\sum_{\frac{C^*\logg T}{\log
T}<|\gamma_n-\gamma_m|\le1}\frac{1}{\gamma_n-\gamma_m}=\mathstrut&\sum_{0\le
k\le
N}\frac{2\gamma_n-\gamma_{m_2+k}-\gamma_{m_1-k}}{(\gamma_n-\gamma_{m_2+k})(\gamma_n-\gamma_{m_1-k})}+O(\log
T)\\
=\mathstrut&O\left(M(n)\sum_{k=1}^{\infty}\frac{\logg T}{\left(\frac{k\logg
T}{\log
T}\right)^2}\right)+O(\log T)\\
=\mathstrut&O(\log T).\end{aligned}$$ Thus, we complete the proof of Proposition \[p:logdea\].

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[^1]: Research of the first author supported by the American Institute of Mathematics, and the National Science Foundation. Research of the second author supported by the Korea Research Foundation Grant funded by the Korean Government(MOEHRD, Basic Research Promotion Fund)(KRF-2008-313-C00009).
