
Section2Proof of Theorem 1.1.2

Theorem 1.1.2 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

$$M_{\gamma_c}=\sum_{0< |\gamma-\gamma_c|\le X(\gamma_c)} \frac{1}{\gamma-\gamma'},\quad\quad\tag{2.1}$$

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 [6], we expect that $M_{\gamma_c}$ should be large if and only if $\beta'-\frac12$ is small. And just like in [6], 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

$$\frac{\zeta'}{\zeta}(s) = \sum_{|\gamma-t|< X(T)} \frac{1}{s-\rho} + O(\log T).\quad\quad\tag{2.2}$$

On RH, with $X(t)=1/\logg T$ the above holds for all $t$ [15]. Using this, instead of our (2.3) below, leads to a weaker version of Theorem 1.1.2, where the $\logg T$ in the denominator of (1.1.2) is replaced by $\log T$.

We prove the following strengthening of (2.2), but only near almost all $\gamma$.

The proof of Proposition 2.1 is in Section 3.2.

Subsection2.1Restricting to zeros with special properties

We begin the proof of Theorem 1.1.2. The lemmas in this section show that, in the context of the proof of Theorem 1.1.2, we only have to deal with zeros that are well spaced.

Suppose, for the purposes of contradiction, that there exists $\epsilon>0$ so that

$$\liminf_{T\to\infty}\frac{\#\{\gamma_n\le T:\gamma_{n+1}-\gamma_n\le\epsilon/\log\gamma_n\}}{T\log T/\logg T}=0.\quad\quad\tag{2.1.1}$$

Then, we can find a sequence $\langle T_l\rangle$ such that $T_1$ is sufficiently large, $T_l\to\infty$ and

$$\#\{\gamma_n\le T_l:\gamma_{n+1}-\gamma_n\le\epsilon/\log\gamma_n\}=o\left(T_l\log T_l/\logg T_l\right)\quad\quad\tag{2.1.2}$$

as $l\to\infty$. We set

$$T=T_l.\quad\quad\tag{2.1.3}$$

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

Proof

For each $m=\pm1,\pm2,\ldots$, let

$$A_m=\{\gamma_n< T:|\gamma_{n+m}-\gamma_{n+m-1}|\geqslant\frac{\epsilon}{2\log\gamma_n}\}.\quad\quad\tag{2.1.5}$$

Here, we exclude the case $n+m\le1$. By assumption (2.1.1) have

$$\#(A_m)=\frac{T}{2\pi}\log T+o\left(\frac{T\log T}{\logg T}\right)\quad\quad\tag{2.1.6}$$

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\quad\quad\tag{2.1.7}

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

$$N(\gamma,l_1,l_2)= N\left(\gamma+\frac{l_2C^*\logg\gamma}{\log\gamma}\right) -N\left(\gamma+\frac{l_1C^*\logg\gamma}{\log\gamma}\right) -\frac{(l_2-l_1)C^*\logg\gamma}{2\pi}\quad\quad\tag{2.1.8}$$

Using an argument in [5], we get the following.

The proof of Proposition 2.1.2 is in Section 3.1.

Subsection2.2Lower 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:

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 [15], Theorem 9.6(A):

$$\frac{\zeta'}{\zeta}(s)=-\frac{1}{2}\log t+O(1)+\sum_{\rho}\left(\frac{1}{s-\rho}-\frac{1}{\rho}\right),\quad\quad\tag{2.2.4}$$

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 (2.2.4) we have

$$\frac{1}{2}\log\gamma'+O(1) = \frac{\beta'-\frac{1}{2}} {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2} + \sum_{\gamma\not=\gamma_c} \frac{\beta'-\frac{1}{2}} {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma)^2}.\quad\quad\tag{2.2.5}$$

There are three cases to consider.

Case 1. $\beta'-1/2>|\gamma'-\gamma_c|$.

Then, by (2.2.5), we get

$$\frac{1}{2}\log\gamma'\geqslant\frac{1}{2\left(\beta'-\frac{1}{2}\right)}.\quad\quad\tag{2.2.6}$$

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 (2.2.5), (2.2.1), and (2.2.2), 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\quad\quad\tag{2.2.7}

and so again we have

$$\beta'-\frac{1}{2}\gg\frac{1}{\log\gamma'}.\quad\quad\tag{2.2.8}$$

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 (2.2.5), (2.2.1), and (2.2.2), as in Case 2, we get

$$\frac{1}{2}\log\gamma'\geqslant\frac{\beta'-\frac{1}{2}}{2(\gamma'-\gamma_c)^2}\quad\quad\tag{2.2.9}$$ $$\frac{1}{2}\log\gamma'\ll\frac{\beta'-\frac{1}{2}}{(\gamma'-\gamma_c)^2}+\left(\beta'-\frac{1}{2}\right)\log^2\gamma'\\ \ll\frac{\beta'-\frac{1}{2}}{(\gamma'-\gamma_c)^2}.\quad\quad\tag{2.2.10}$$

Thus we have

$$(\gamma'-\gamma_c)^2\log\gamma'\ll \beta'-\frac{1}{2}\ll(\gamma'-\gamma_c)^2\log\gamma'.\quad\quad\tag{2.2.11}$$

Here the implied constants depend only on $\epsilon$. By (2.2.11) and the conditions of Case 3 we have

$$\frac{\gamma_c-\gamma'}{\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2} -\frac{1}{\gamma_c-\gamma'}=O(\log\gamma').\quad\quad\tag{2.2.12}$$

Now take the imaginary part of (2.2.3) to get

$$\sum_{0< |\gamma-\gamma_c| \le \frac{C^*\logg\gamma_c}{\log\gamma_c}} \frac{\gamma-\gamma'}{\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma)^2} +\frac{\gamma_c-\gamma'} {\left(\beta'-\frac{1}{2}\right)^2+(\gamma'-\gamma_c)^2} = O\left(\log\gamma'\right).\quad\quad\tag{2.2.13}$$

Finally, by (2.2.1) we have

where

$$M_{\gamma_c}=\sum_{0< |\gamma-\gamma_c|\le \frac{C^*\logg\gamma_c}{\log\gamma_c}}\frac{1}{\gamma-\gamma'}.\quad\quad\tag{2.2.16}$$

By combining (2.2.12), (2.2.13), (2.2.14), and (2.2.11), we have

$$O(\log\gamma')=M_{\gamma_c}+\frac{1}{\gamma_c-\gamma'}=M_{\gamma_c} + A_{\gamma_c} \sqrt{\frac{\log\gamma'}{\beta'-\frac{1}{2}}},\quad\quad\tag{2.2.17}$$

where $1\ll A_{\gamma_c}\ll 1$, with the implied constants depending only on $\epsilon$.

Let $\nu$ be a positive number. Suppose that

$$\left(\beta'-\frac{1}{2}\right)\log\gamma'\le\nu.\quad\quad\tag{2.2.18}$$

Then, for sufficiently small $\nu$, we see that only Case 3 is possible for sufficiently large $\gamma'$, namely we have

$$M_{\gamma_c}+ A_{\gamma_c} \sqrt{\frac{\log\gamma'}{\beta'-\frac{1}{2}}} =O(\log \gamma').\quad\quad\tag{2.2.19}$$

By this, the assumption in Theorem 1.1.2, 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

$$\frac{e^{-C(\nu)}\log^{2k}T}{\nu^{k}}(1+O(\sqrt{\nu}))^{2k}\frac{T}{2\pi}\log T\ll\sum_{\substack{\frac{T}{\log T}\le\gamma'\le T\\ \left(\beta'-\frac{1}{2}\right)\log\gamma'\le\nu\\ \gamma_c\in Z(T)} } |M_{\gamma_c}|^{2k}.\quad\quad\tag{2.2.20}$$

In the next section we describe upper bounds for the moments of $M_{\gamma_c}$. This will contradict (2.2.20) and complete the proof of Theorem 1.1.2.

Subsection2.3Bounding the moments of $M_{\gamma_c}$

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.

Proof

By the assumptions we have

Proof

By [15], Theorem 14.20,

The assumptions on the zero spacings give the claimed bound on the terms involving the zeros.

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

By Lemma 2.3.1 and Lemma 2.3.2, with $A=A_\epsilon$ a constant depending only on $\epsilon$, which may be different in each inequality, we have

\begin{align} |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}\\ \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}\\ \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},\quad\quad\quad\quad\quad\quad \quad \tag{2.3.9} \end{align}

where $x=T^{1/100k}$, for $|t-\gamma'|\le \delta(\epsilon)/\log\gamma'$, provided $\gamma_c$ satisfies (2.2.1)(2.2.3). That is, provided $\gamma_c\in Z(T)$.

Integrating inequality (2.3.8) over the set

$$\left\{T/{\log T} < t < T\ :\ |t-\gamma_c|< \delta(\epsilon)/\log T \text{ for some } \gamma_c\in Z(T) \right\}\quad\quad\tag{2.3.10}$$

and then using Lemma 2.3.3 we get

\begin{align}\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\\ \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}\\ \ll \mathstrut& A^{2k} k^{2k} T\log^{2k}T.\quad\quad\quad\quad\quad\quad \quad \tag{2.3.11} \end{align}

The last step used $\Lambda_x(p)\le \Lambda(p)$ and the fact that

$$\sum_{p\le x}\frac{\Lambda(p)^2}{p} \ll \log^2 x,\quad\quad\tag{2.3.12}$$

which is a weak form of the prime number theorem.

Rearranging the above inequality and combining with (2.2.20), we have

$$\frac{e^{-C(\nu)}}{\nu^k}(1+O(\sqrt{\mathstrut\nu}))^{2k}T\log^{2k+1}T \ll A^{2k} k^{2k} T\log^{2k+1}T,\quad\quad\tag{2.3.13}$$

which rearranges to give

$$\left(1+O(\sqrt{\mathstrut\nu})\right)^{2k} \ll A^{2k} k^{2k} \nu^k e^{C(\nu)}.\quad\quad\tag{2.3.14}$$

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 1.1.2.