An Approximation to Riemann Hypothesis

1. Introduction

Riemann zeta-function $\zeta \left(s\right)$ is originally defined as

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}},\mathrm{for}\mathrm{Re}s>1.$$

and it also can be expressed as the product form

$$\zeta \left(s\right)=\prod _p{\displaystyle \frac{1}{1-1/p^s}},\mathrm{for}\mathrm{Re}s>1.$$

This formula is called Euler’s product formula, which indicates the relation between $\zeta \left(s\right)$ and prime numbers. About $\zeta \left(s\right)$ there is a well-known Riemann hypothesis, states that all the non-trivial zeros of $\zeta \left(s\right)$ are on the critical line $\mathrm{Re}s=1/2$. The researches on the conjecture are no doubt a most time-consuming one in mathematics, refer to see the survey paper [3].

The so-called trivial zeros of $\zeta \left(s\right)$ are $s=-2,-4,\dots$, and nontrivial zeros of $\zeta \left(s\right)$ are known all in the critical strip $0\le \mathrm{Re}s\le 1$.

Denote by $N\left(T\right)$ the number of zeros of $\zeta (\sigma +it)$ in the region $0\le \sigma \le 1$, $0\le t\le T$, and by $N_0\left(T\right)$ the number of zeros on the critical line $\sigma =1/2$, $0\le t\le T$. Riemann hypothesis is that

$$N_0\left(T\right)=N\left(T\right).$$

For $N\left(T\right)$, it is known that

$$N\left(T\right)={\displaystyle \frac{T}{2\pi }}log{\displaystyle \frac{T}{2\pi }}-{\displaystyle \frac{T}{2\pi }}+O(logT).$$

(1.1)

And for $N_0\left(T\right)$, Hardy firstly shown that there are infinity many zeros on the critical line, and then he and Littlewood [5] and Selberg [8] proved that

$$\kappa ={\displaystyle \frac{N_0\left(T\right)}{N\left(T\right)}}>0.$$

Levinson [6] proved

$$\kappa \ge {\displaystyle \frac{1}{3}}.$$

and then this result has been improved successively. Conrey [2], Feng [4] proved respectively

$$\kappa \ge 0.407,\kappa \ge 0.412.$$

In this paper, we will prove that

Theorem 1.1.

$$N\left(T\right)=N_0\left(T\right)+\mathcal{E}.$$

(1.2)

where $\mathcal{E}\ll T^{1/2}{(logT)}^3$.

The main arguments in this paper are based the papers [1,6,7], but instead of using Riemann-Siegel formula, it will be applied an auxiliary function $\omega (s,T_1,T_2)$ defined in Lemma 2.1, which will play a role of mollifier and ferry, it firstly used in [7] but here with a small modification.

2. Some Lemmas

In the following, it will be used an auxiliary function $\omega (s,T_1,T_2)$, which defined as following.

Suppose that $T\le T_1\le T_2\le 2T$, $\lambda =T^{1-ϵ}$, for any $ϵ>0$, $s_1=\lambda +c+iu$,$(c\ge 0)$, $s=v+it$, define

$$g\left(w\right)=e^{\lambda }\Gamma \left(w\right){\lambda }^{-w},$$

and

$$\omega (s,T_1,T_2)={\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}g(s_1-s)du.$$

Lemma 2.1.

Let $\sigma =\lambda +c-v$, $\eta =min\left\{|x-t||x\in [T_1,T_2]\right\}$, there is

$$|\omega (s,T_1,T_2)|\ll exp(({(c-v)}^2-{\eta }^2)/2\sigma )$$

(2.2)

Suppose $\Delta \ge {\Delta }_0(={(2\beta \lambda logT)}^{1/2},\beta >0)$, if $t\in [T_1+\Delta ,T_2-\Delta ]$, then

$$|\omega (c+it,T_1,T_2)-1|\ll T^{-\beta }.$$

(2.3)

And if $t\le T_1-\Delta$, or $t\ge T_2+\Delta$, then

$$|\omega (c+it,T_1,T_2)|\ll T^{-\beta }.$$

(2.4)

If $|t-u|=o\left({\sigma }^{2/3}\right)$, then

$$argg(s_1-s)={\displaystyle \frac{t-u}{2\sigma }}-{\displaystyle \frac{{(t-u)}^3}{6{\sigma }^2}}-{\displaystyle \frac{(c-v)(t-u)}{\sigma }}+O\left({\displaystyle \frac{1}{\sigma }}\right).$$

(2.5)

Proof. 

By Stirling’s formula, it has

$$\begin{array}{cc}\hfill \mathrm{Re}(log\Gamma (\sigma +(u-t)i\left)\right)=& \left(\sigma -{\displaystyle \frac{1}{2}}\right){\displaystyle \frac{log({\sigma }^2+{(u-t)}^2)}{2}}-\sigma +{\displaystyle \frac{1}{2}}log\left(2\pi \right)\hfill \\ \hfill & -(u-t)arctan\left({\displaystyle \frac{u-t}{\sigma }}\right)+O\left({\displaystyle \frac{1}{\sigma }}\right)\hfill \end{array}$$

And

$$\begin{array}{cc}\hfill & \mathrm{Re}(log\left(\Gamma (\sigma +(u-t)i)\right))+\mathrm{Re}(log\left({\lambda }^{-(\sigma +(u-t\left)i\right)}\right))+\lambda \hfill \\ \hfill & ={\displaystyle \frac{{(c-v)}^2}{2\sigma }}-{\displaystyle \frac{{(u-t)}^2}{4{\sigma }^2}}-{\displaystyle \frac{{(u-t)}^2}{2\sigma }}-{\displaystyle \frac{1}{2}}log\sigma +{\displaystyle \frac{1}{2}}log\left(2\pi \right)+O\left({\displaystyle \frac{1}{\sigma }}\right)\hfill \end{array}$$

Hence,

$$|\omega (s,T_1,T_2)|\le e^{{(c-v)}^2/2\sigma }{\displaystyle \frac{1}{\sqrt{2\pi \sigma }}}{\int }_{T_1}^{T_2}{e}^{-{(u-t)}^2/2\sigma }du\ll e^{({(c-v)}^2-{\eta }^2)/2\sigma }.$$

Besides, it is familiar that

$$e^{-\lambda }={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){\lambda }^{s-s_1}ds.$$

Hence,

$$1-\omega (c+it,T_1,T_2)=R_1+R_2,$$

where

$$R_1={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_2}^{\infty }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du,$$

$$R_2={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{-\infty }^{T_1}\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du,$$

Hence, if $t\in [T_1+\Delta ,T_2-\Delta ]$, then

$$\begin{array}{cc}\hfill \left|R_1\right|& \ll {\int }_{T_2}^{\infty }\left|e^{\lambda }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}\right|du\hfill \\ \hfill & \ll {\lambda }^{-1/2}{\int }_{T_2}^{\infty }exp(-{(u-t)}^2/2\lambda )du\hfill \\ \hfill & \ll T^{-\beta }.\hfill \end{array}$$

and similarly,

$$\left|R_2\right|\ll {\lambda }^{-1/2}{\int }_{-\infty }^{T_1}exp(-{(u-t)}^2/2\lambda )du\ll T^{-\beta }.$$

if $t\le T_1-\Delta$, or $t\ge T_2+\Delta$, then

$$\left|\omega (c+it,T_1,T_2)\right|\ll {\lambda }^{-1/2}{\int }_{T_1}^{T_2}exp(-{(u-t)}^2/2\lambda )du\ll T^{-\beta }.$$

If $|t-u|=o\left({\sigma }^{2/3}\right)$, then

$$\begin{array}{cc}\hfill \mathrm{Im}(log\left(\Gamma ({\mathrm{s}}_1-\mathrm{s}){\lambda }^{\mathrm{s}-{\mathrm{s}}_1}\right))=& {\displaystyle \frac{t-u}{2\sigma }}+{\displaystyle \frac{{(t-u)}^3}{3{\sigma }^2}}-{\displaystyle \frac{{(t-u)}^3}{2{\sigma }^2}}\hfill \\ \hfill & -{\displaystyle \frac{(c-v)(t-u)}{\sigma }}+O\left({\displaystyle \frac{1}{\sigma }}\right)\hfill \\ \hfill =& {\displaystyle \frac{t-u}{2\sigma }}-{\displaystyle \frac{{(t-u)}^3}{6{\sigma }^2}}-{\displaystyle \frac{(c-v)(t-u)}{\sigma }}+O\left({\displaystyle \frac{1}{\sigma }}\right).\hfill \end{array}$$

□

Lemma 2.2.

Let $L=log(T/2\pi )$, $G_0\left(s\right)=\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)/L$, and $0<a\le 1/2$, $1<b<\beta -1$, $u\in [T_1,T_2]$, $s_1=\lambda +c+iu$, then

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2\pi i}}{\int }_{a+(u-\Delta )i}^{a+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi i}}{\int }_{b+(u-\Delta )i}^{b+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds+O(b/T)\hfill \end{array}$$

(2.6)

Proof. 

Let $\mathcal{B}$ be the rectangle with vertices $a+(u-\Delta )i$, $a+(u+\Delta )i$, $b+(u-\Delta )i$ and $b+(u+\Delta )i$. Take the integral $\underset{\mathcal{B}}{\int }g(s_1-s)G\left(s\right)G(2a-s)ds$, with the residue theorem, it has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2\pi i}}{\int }_{b+(u-\Delta )i}^{b+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi i}}{\int }_{a+(u-\Delta )i}^{a+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds\hfill \\ \hfill & +{\displaystyle \frac{1}{2\pi i}}{\int }_{a+(u+\Delta )i}^{b+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds\hfill \\ \hfill & +{\displaystyle \frac{1}{2\pi i}}{\int }_{b+(u-\Delta )i}^{a+(u-\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds.\hfill \end{array}$$

On the upper side and the lower side, as $\left|\mathrm{Im}\right(s_1-s\left)\right|=\Delta$, by Lemma 2.1, it has

$$|g(s_1-s)|\ll 1/T^{\beta }.$$

Besides, by the functional equation, it is easy to know

$$|G_0(v+(u\pm \Delta )i)G_0(2a-(v+(u\pm \Delta )i))|\ll T^b,(a\le v\le b).$$

Hence, the two integrals on the upper side and the lower side of $\mathcal{B}$

$${\displaystyle \frac{1}{2\pi i}}{\int }_{a+(u\pm \Delta )i}^{b+(u\pm \Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds\ll b/T,$$

and the Lemma is followed.

□

Lemma 2.3.

Suppose that $0<\alpha <\beta -2$. For $r>0$, let

$$J\left(r\right)={\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{u-\Delta }^{u+\Delta }{\left({\displaystyle \frac{t}{2\pi }}\right)}^{\alpha }{e}^{-{(t-u)}^2/2\sigma }exp\left(itlog\left({\displaystyle \frac{t}{re}}\right)\right)dtdu$$

Then for $T_1-\Delta \le r\le T_2+\Delta$,

$$J\left(r\right)\doteq {\left({\displaystyle \frac{r}{2\pi }}\right)}^{\alpha }{r}^{1/2}{\sigma }^{1/2}{e}^{\pi i/4}.$$

(2.7)

And if $r<T_1-\Delta$, or $r>T_2+\Delta$, and $\Delta \ge {\Delta }_0{T}^{ϵ}$, then

$$J\left(r\right)=O\left(1\right).$$

(2.8)

Proof. 

Denote by

$$F\left(u\right)={\int }_{u-\Delta }^{u+\Delta }{\left({\displaystyle \frac{t}{2\pi }}\right)}^{\alpha }{e}^{-{(t-u)}^2/2\sigma }exp\left(itlog\left({\displaystyle \frac{t}{re}}\right)\right)dt$$

And let $t=u(1+x)$,then

$$F\left(u\right)={\left({\displaystyle \frac{u}{2\pi }}\right)}^{\alpha }u{e}^{iu\rho }{F}_1\left(u\right),$$

where $\rho =log(u/r)-1$,

$$F_1\left(u\right)={\int }_{-\Delta /u}^{\Delta /u}exp\left(A\left(x\right)+B\left(x\right)i\right)dx,$$

$$A\left(x\right)=\alpha log(1+x)-{\left(ux\right)}^2/2\sigma ,B\left(x\right)=u(1+x)log(1+x)+ux\rho .$$

By Gauss’s integration, it is easy to follow

$$F_1\left(u\right)\doteq {\displaystyle \frac{1}{u}}{\displaystyle \frac{2\pi \sigma }{\sqrt{1+\alpha \sigma /u^2-i\sigma /u}}}exp\left(-{\displaystyle \frac{{(1+\rho -i\alpha /u)}^{2}u}{2(u/\sigma +\alpha /u-i)}}\right)$$

and

$$F\left(u\right)\doteq {\left({\displaystyle \frac{u}{2\pi }}\right)}^{\alpha }{e}^{iu\rho }{\displaystyle \frac{\sqrt{2\pi \sigma }}{\sqrt{1+\alpha \sigma /u^2-i\sigma /u}}}exp\left(-{\displaystyle \frac{{(log(u/r)-i\alpha /u)}^2\sigma }{2(1+\alpha \sigma /u^2-i\sigma /u)}}\right)$$

If $r\in [T_1-\Delta ,T_2+\Delta ]$, let $u=r(1+x)$, then

$$F\left(u\right)\doteq {\left({\displaystyle \frac{r}{2\pi }}\right)}^{\alpha }\sqrt{2\pi \sigma }exp\left(-{\displaystyle \frac{x^2(\sigma -ri)}{2}}\right){(1+x)}^{\alpha }$$

and

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2\pi }}{\int }_{T_1/r-1}^{T_2/r-1}F\left(u\right)rdx& \doteq {\displaystyle \frac{r}{2\pi }}{\left({\displaystyle \frac{r}{2\pi }}\right)}^{\alpha }{\displaystyle \frac{\sqrt{2\pi \sigma }·\sqrt{\pi }}{\sqrt{(\sigma +\alpha -ri)/2}}}exp\left({\displaystyle \frac{{\alpha }^2(\sigma +\alpha +ri)}{2{(\sigma +\alpha )}^2+r^2)}}\right)\hfill \\ \hfill & \doteq {\left({\displaystyle \frac{r}{2\pi }}\right)}^{\alpha }{r}^{1/2}\sqrt{\sigma }{e}^{\pi i/4}\hfill \end{array}$$

If $r<T_1-\Delta$, or $r>T_2+\Delta$, and if $\Delta \ge {\Delta }_0{T}^{ϵ}$, then

$$\begin{array}{cc}\hfill \left|F\right(u\left)\right|& \ll T^{\alpha +1/2}exp\left(-{\left({\displaystyle \frac{\Delta }{T}}\right)}^2{\displaystyle \frac{\sigma }{2}}\right)\hfill \\ \hfill & \ll T^{\alpha +1/2-\beta }\ll O(1/T).\hfill \end{array}$$

□

3. The Proof of Theorem 1.1

Proof. 

Let $h\left(s\right)={\pi }^{-s/2}\Gamma (s/2)$, then the functional equation of $\zeta \left(s\right)$ can be written as

$$h\left(s\right)\zeta \left(s\right)=h(1-s)\zeta (1-s)$$

(3.1)

By Stirling’s formula, it has

$$logh\left(s\right)={\displaystyle \frac{1}{2}}(s-1)log{\displaystyle \frac{s}{2\pi }}-{\displaystyle \frac{s}{2}}+C_0+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.2)

Let $f\left(s\right)=logh\left(s\right)$, then

$$f^{\prime }\left(s\right)={\displaystyle \frac{h^{\prime }\left(s\right)}{h\left(s\right)}}={\displaystyle \frac{1}{2}}log{\displaystyle \frac{s}{2\pi }}+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.3)

and for larger t

$$f^{\prime }\left(s\right)+f^{\prime }(1-s)=log{\displaystyle \frac{t}{2\pi }}+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.4)

Taking logarithm of equation (3.1), and then derivative, it follows

$$h\left(s\right)\zeta \left(s\right)(f^{\prime }\left(s\right)+f^{\prime }(1-s))=-h\left(s\right){\zeta }^{\prime }\left(s\right)-h(1-s){\zeta }^{\prime }(1-s)$$

(3.5)

We note that the right side of (3.5) is a sum of two conjugative complex numbers as $s=1/2+it$, so the zeros of the right side of (3.5) occur if and only if

$$arg\left(h\left(s\right){\zeta }^{\prime }\left(s\right)\right)\equiv \pi /2mod\pi$$

(3.6)

On the left side of (3.5), clearly, $h\left(s\right)$ is never zero, and by (3.4), so these zeros are just the zeros of $\zeta (1/2+it)$.

Moreover, let $\chi \left(s\right)=h(1-s)/h\left(s\right)$, then $\zeta \left(s\right)=\chi \left(s\right)\zeta (1-s)$, and

$${\zeta }^{\prime }\left(s\right)=-\chi \left(s\right)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta (1-s)+{\zeta }^{\prime }(1-s)\}$$

(3.7)

By (3.6), the zeros of $\zeta (1/2+it)$ are the ones

$$arg\left(h(1-s)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta (1-s)+{\zeta }^{\prime }(1-s)\}\right)\equiv \pi /2mod\pi$$

on $\sigma =1/2$, equivalently,

$$arg\left(h\left(s\right)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)\}\right)\equiv \pi /2mod\pi$$

(3.8)

on $\sigma =1/2$. Write $\mathcal{L}\left(s\right)=f^{\prime }\left(s\right)+f^{\prime }(1-s)$, and denote by

$$G\left(s\right)=\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)/\mathcal{L}\left(s\right)$$

(3.9)

The investigation above means

$$N_0\left(T\right)={\displaystyle \frac{1}{\pi }}{\Delta }_0^{T}arg\left(hG(1/2+it)\right)$$

(3.10)

By(3.2), it can be known that

$${\Delta }_0^{T}arg\left(h(1/2+it)\right)={\displaystyle \frac{T}{2}}log{\displaystyle \frac{T}{2\pi }}-{\displaystyle \frac{T}{2}}+O(logT)$$

(3.11)

So, the main task to determine $N_0\left(T\right)$ is to calculate ${\Delta }_0^{T}arg\left(G(1/2+it)\right)$.

Let $L=log(T/2\pi )$, $U\le T$, and let D be the rectangle with the vertices $1/2+iT$, $c+iT$, $c+i(T+U)$, $1/2+i(T+U)$,$(c\ge 3)$. First of all, we might as well assume there are no zeros of $G\left(s\right)$ on the boundary of D, then by the principle of argument, the change of $argG\left(s\right)$ around D is equal to $2\pi$ times $N_G\left(D\right)$, the number of zeros of $G\left(s\right)$ in D.

On the right side of D

$$|G(c+it)-1|\le \sum _{n\ge 2}{n}^{-c}+O(1/L)\le 1/3$$

so, $argG\left(s\right)$ change less than $\pi$. On the lower side and the upper side of D, by a known result [9, $\S 9.4$], a extension of Jessen’s theorem, taking account on the order of $G\left(s\right)$, we can know that $argG\left(s\right)=O\left(L\right)$ as $0<\sigma \le 3$, and $argG(\sigma +it)=O\left(2^{-\sigma }\right)$ as $\sigma \ge 3$, hence, for any $0\le b\le c$, it has

$${\int }_b^{c}argG(\sigma +iT)d\sigma ,{\int }_b^{c}argG(\sigma +i(T+U))d\sigma \ll O\left(L\right)$$

(3.12)

So,

$${\Delta }_T^{T+U}arg\left(G(1/2+it)\right)=-2\pi N_G\left(D\right)+O(logT)$$

(3.13)

Now the work is turned into to evaluate $N_G\left(D\right)$.

Let $1/2-a=O(1/L)$, and $\mathcal{C}$ be the rectangle with vertices $a+iT$, $c+iT$, $c+i(T+U)$, $a+i(T+U)$. Taking the integral $\underset{\mathcal{C}}{\int }logG\left(s\right)ds$, by the Littlewood’s Lemma [9, $\S 9.9$], it has

$$\begin{array}{cc}\hfill & {\int }_T^{T+U}log\left|G(a+it)\right|dt-{\int }_T^{T+U}log\left|G(c+it)\right|dt+{\int }_a^{c}argG(\sigma +i(T+U))d\sigma \hfill \\ \hfill & -{\int }_a^{c}argG(\sigma +iT)d\sigma =2\pi \sum \mathrm{dist}\hfill \end{array}$$

(3.14)

where $\sum \mathrm{dist}$ is the sum of the distances of the zeros of $G\left(s\right)$ from the left.

By (3.9), it is easy to know

$${\int }_T^{T+U}logG(c+it)dt={\int }_T^{T+U}log\zeta (c+it)dt+O(1/L)$$

and it is familiar that

$$log\zeta \left(s\right)=\sum _n{\displaystyle \frac{-\Lambda \left(n\right)}{n^{s}logn}}$$

So

$${\int }_T^{T+U}log\left|G(c+it)\right|dt\ll 1.$$

With (3.12), the rest is to calculate the first integral of (3.14).

By the concavity of logarithm, it has

$$\begin{array}{cc}\hfill {\int }_T^{T+U}log\left|G(a+it)\right|dt& ={\displaystyle \frac{1}{2}}{\int }_T^{T+U}log{\left|G(a+it)\right|}^{2}dt\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}Ulog\left({\displaystyle \frac{1}{U}}{\int }_T^{T+U}{\left|G(a+it)\right|}^{2}dt\right)\hfill \end{array}$$

(3.15)

At first, we simplify $G\left(s\right)$ as

$$G_0\left(s\right)=\zeta \left(s\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(s\right)}{L}}.$$

Then

$$G\left(s\right)=G_0\left(s\right)+E\left(s\right).$$

$$E\left(s\right)=\left({\displaystyle \frac{1}{\mathcal{L}\left(s\right)}}-{\displaystyle \frac{1}{L}}\right){\zeta }^{\prime }\left(s\right)\ll {\displaystyle \frac{1}{L^3}}{\zeta }^{\prime }\left(s\right).$$

And

$$\begin{array}{cc}\hfill {\int }_{T_1}^{T_2}{\left|G(a+it)\right|}^{2}dt=& {\int }_{T_1}^{T_2}{\left|G_0(a+it)\right|}^{2}dt+2\mathrm{Re}{\int }_{T_1}^{T_2}{G}_0(a+it)E(a-it)dt\hfill \\ \hfill & +{\int }_{T_1}^{T_2}{\left|E(a+it)\right|}^{2}dt\hfill \end{array}$$

(3.16)

By Cauchy’s inequality

$${\int }_{T_1}^{T_2}{G}_0(a+it)E(a-it)dt\le {\left({\int }_{T_1}^{T_2}|G_0{(a+it)|}^{2}dt{\int }_{T_1}^{T_2}{\left|E(a+it)\right|}^{2}dt\right)}^{1/2}$$

The third integral in the right side of (3.16) is much smaller than the first one, which will be actually calculated later, hence

$${\int }_{T_1}^{T_2}{\left|G(a+it)\right|}^{2}dt=(1+ϵ){\int }_{T_1}^{T_2}{\left|G_0(a+it)\right|}^{2}dt.$$

Let

$${\omega }_1\left(s\right)={\displaystyle \frac{1}{2\pi i}}\underset{|u-t|\le \Delta }{\int }g(s_1-s)d{s}_1.$$

(3.17)

From Lemma 2.1, it is known that ${\omega }_1\left(s\right)$ is the domination of $\omega (s,T_1,T_2)$, and which is a positive real number apart from a small error term.

Moreover, let

$$\varphi \left(s\right)={\omega }_1^{1/2}\left(s\right).$$

(3.18)

Then it has

$$arg\varphi (v+T_{1}i)=o\left(1\right),arg\varphi (v+T_{2}i)=o\left(1\right).$$

(3.19)

And

$${\int }_a^{c}arg\varphi (v+T_{1}i)dv\le o\left(c\right),{\int }_a^{c}arg\varphi (v+T_{2}i)dv\le o\left(c\right).$$

(3.20)

Moreover, by Lemma 2.1, there is

$${\int }_{T_1}^{T_2}log\left|\varphi (c+it)\right|dt\ll T^{-b}.$$

(3.21)

(3.19) and (3.20) indicate that function $\varphi \left(s\right)$ may be used as a mollifier.

Let

$$\mathcal{G}\left(s\right)=G_0\left(s\right)\varphi \left(s\right).$$

(3.22)

In the following we will replace $G\left(s\right)$ by $\mathcal{G}\left(s\right)$, and let

$$\mathcal{I}={\int }_{T_1}^{T_2}{\left|\mathcal{G}\left(s\right)\right|}^{2}dt$$

Then,

$$\begin{array}{cc}\hfill \mathcal{I}& ={\int }_{T_1}^{T_2}{\left|G_0\left(s\right)\right|}^2{\omega }_1\left(s\right)dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{t-\Delta }^{t+\Delta }g(s_1-s)du{\left|G_0(a+it)\right|}^{2}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{u-\Delta }^{u+\Delta }g(s_1-s){\left|G_0(a+it)\right|}^{2}dtdu+R_1+R_2-R_3-R_4\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill R_1=& {\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_1+\Delta }{\int }_{u-\Delta }^{T_1}g(s_1-s)du{\left|G_0(a+it)\right|}^{2}dt,\hfill \\ \hfill R_2=& {\displaystyle \frac{1}{2\pi }}{\int }_{T_2-\Delta }^{T_2}{\int }_{u+\Delta }^{T_1}g(s_1-s)du{\left|G_0(a+it)\right|}^{2}dt,\hfill \\ \hfill R_3=& {\displaystyle \frac{1}{2\pi }}{\int }_{T_1-\Delta }^{T_1}{\int }_{T_1}^{u+\Delta }g(s_1-s)du{\left|G_0(a+it)\right|}^{2}dt,\hfill \\ \hfill R_4=& {\displaystyle \frac{1}{2\pi }}{\int }_{T_2}^{T_2+\Delta }{\int }_{u-\Delta }^{T_2}g(s_1-s)du{\left|G_0(a+it)\right|}^{2}dt.\hfill \end{array}$$

By the mean-value theorems (cf. [9, Ch.7]), it has

$$R_1\ll {\int }_{T_1}^{T_1+\Delta }{\left|G_0(a+it)\right|}^{2}dt\ll \Delta L,$$

similarly,

$$R_i\ll \Delta L,i=2,3,4.$$

Then, by Lemma 2.2, it has

$$\mathcal{I}={\displaystyle \frac{1}{2\pi i}}{\int }_{T_1}^{T_2}{\int }_{b+(u-\Delta )i}^{b+(u+\Delta )i}g(s_1-s)G_0\left(s\right)G_0(2a-s)ds+O(\Delta L)$$

Moreover, by the functional equation of $\zeta \left(z\right)$, it has

$$\zeta (2a-(b+it\left)\right)=\chi (2a-(b+it\left)\right)\zeta (1+b-2a-it),$$

$${\zeta }^{\prime }(2a-(b+it))=-\chi (2a-(b+it))(L\zeta (1+b-2a+it)+{\zeta }^{\prime }(1+b-2a+it))$$

and

$$G_0(2a-(b+it))=-\chi (2a-(b+it)){\displaystyle \frac{{\zeta }^{\prime }(1+b-2a+it)}{L}}$$

and it is easy to know

$$\chi (2a-(b+it))=t^{1/2+b-2a}exp\left(-{\displaystyle \frac{\pi i}{4}}+itlog\left({\displaystyle \frac{t}{2\pi e}}\right)\right)$$

Hence,

$$\mathcal{I}=I_1+I_2+O(\Delta L).$$

where

$$I_1=-{\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{u-\Delta }^{u+\Delta }g(s_1-s)\chi (2a-(b+it))\zeta (b+it){\displaystyle \frac{{\zeta }^{\prime }(1+b-2a+it)}{L}}dtdu$$

$$I_2=-{\displaystyle \frac{1}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{u-\Delta }^{u+\Delta }g(s_1-s)\chi (2a-(b+it)){\displaystyle \frac{{\zeta }^{\prime }(b+it)}{L}}{\displaystyle \frac{{\zeta }^{\prime }(1+b-2a+it)}{L}}dtdu$$

Expanding $\zeta \left(s\right)$ and ${\zeta }^{\prime }\left(s\right)$ as Dirichlet’s series, and dividing $I_1$ and $I_2$ into three parts respectively,

$$I_1=I_{1,1}+I_{1,2}+I_{1,3}=\sum _{2\pi xy<T_1-\Delta }+\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }+\sum _{2\pi xy>T_2+\Delta },$$

$$I_2=I_{2,1}+I_{2,2}+I_{2,3}=\sum _{2\pi xy<T_1-\Delta }+\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }+\sum _{2\pi xy>T_2+\Delta }.$$

Then by Lemmas 2.1,∼,3, there are

$$I_{1,1},I_{1,3},I_{2,1},I_{2,3}\ll o\left(1\right).$$

and

$$\begin{array}{cc}\hfill I_{1,2}& ={\displaystyle \frac{2\pi }{L}}\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }{\displaystyle \frac{{\left(xy\right)}^{b^{\prime }}logy}{x^b{y}^{b^{\prime }}}}\hfill \\ \hfill I_{2,2}& ={\displaystyle \frac{-2\pi }{L^2}}\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }{\displaystyle \frac{{\left(xy\right)}^{b^{\prime }}logxlogy}{x^b{y}^{b^{\prime }}}}.\hfill \end{array}$$

where $b^{\prime }=b+1-2a$, and $b=c(>1)$.

i.e.

$$\begin{array}{cc}\hfill I_{1,2}& ={\displaystyle \frac{2\pi }{L}}\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }{x}^{1-2a}logy,\hfill \\ \hfill I_{2,2}& ={\displaystyle \frac{-2\pi }{L^2}}\sum _{T_1-\Delta \le 2\pi xy\le T_2+\Delta }{x}^{1-2a}logxlogy.\hfill \end{array}$$

Let

$$H\left(n\right)={\displaystyle \frac{1}{L}}\sum _{1\le xy\le n}{x}^{1-2a}logy-{\displaystyle \frac{1}{L^2}}\sum _{1\le xy\le n}{x}^{1-2a}logxlogy.$$

It is easy to follow that

$$H\left(n\right)\doteq {\displaystyle \frac{n}{L^2(1-a)}}\left({\displaystyle \frac{n^{1-2a}}{{(1-2a)}^3}}-{\displaystyle \frac{{log}^{2}n}{2(1-2a)}}-{\displaystyle \frac{logn}{{(1-2a)}^2}}-{\displaystyle \frac{1}{{(1-2a)}^3}}\right)$$

Let $x=(1-2a)L$, then

$$H\left(n\right)\doteq {\displaystyle \frac{2nL}{x^3}}(e^x-1-x-x^2/2)\doteq nL/3$$

(3.23)

Obviously, estimation (3.23) is not sufficient for the proof of Theorem 1.1. Nevertheless, there is an alternative way to improve it.

Actually, the argument of Levinson [6] can be extended to differentiate the functional equation of $\zeta \left(s\right)$ to higher order k, and similarly to obtain the functions $G(s,k)$, $k=1,2,...$, and similar results as (3.23), and more sharper.

For examples, for $k=1$,

$$G(s,1)=\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)/L,$$

and

$$H(n,1)={\displaystyle \frac{nL}{x^3}}\{a_1\left(x\right)e^x-b_1\left(x\right)\}\doteq nL/3.$$

(3.24)

where

$$\begin{array}{cc}\hfill a_1\left(x\right)& =2,\hfill \\ \hfill b_1\left(x\right)& =2+2x+x^2.\hfill \end{array}$$

For $k=2$,

$$G(s,2)=\zeta \left(s\right)+4{\zeta }^{\prime }\left(s\right)/L+4{\zeta }^{\prime \prime }\left(s\right)/L^2.$$

and

$$H(n,2)\doteq {\displaystyle \frac{nL}{x^5}}\{a_2\left(x\right)e^x-b_2\left(x\right)\}\doteq nL/5$$

(3.25)

where

$$\begin{array}{cc}\hfill a_2\left(x\right)& =384-192x+48x^2-8x^3+x^4,\hfill \\ \hfill b_2\left(x\right)& =384+192x+48x^2+8x^3+x^4.\hfill \end{array}$$

And for $k=3$,

$$G(s,3)=\zeta \left(s\right)+6{\zeta }^{\prime }\left(s\right)/L+12{\zeta }^{\prime \prime }\left(s\right)/L^2+8{\zeta }^{\prime \prime \prime }\left(s\right)/L^3.$$

and

$$H(n,3)\doteq {\displaystyle \frac{nL}{x^7}}\{a_3\left(x\right)e^x-b_3\left(x\right)\}\doteq nL/7.$$

(3.26)

where

$$\begin{array}{cc}\hfill a_3\left(x\right)& =46080-23040x+5760x^2-960x^3+120x^4-12x^5+x^6,\hfill \\ \hfill b_3\left(x\right)& =46080+23040x+5760x^2+960x^3+120x^4+12x^5+x^6.\hfill \end{array}$$

So, it is predictable that for $k\ge L/2$, there will be

$$H(n,k)\doteq n.$$

A rough proof for this is added in the appendix.

So,

$$\mathcal{I}=U+T^{ϵ},$$

where $ϵ$ is a arbitrary small positive number.

And

$${\int }_T^{T+U}{\left|\mathcal{G}(a+it)\right|}^{2}dt=U+T^{ϵ}.$$

Let $U=T$, by (3.15), it follows

$$\begin{array}{cc}\hfill & {\int }_T^{T+U}log\left|\mathcal{G}(a+it)\right|dt\le {\displaystyle \frac{T}{2}}log\left(1+T^{-1+ϵ}\right)\hfill \\ \hfill & \ll T^{ϵ}.\hfill \end{array}$$

(3.27)

With (3.12), (3.14), (3.19),(3.20) and (3.27), and $(1-2a)L=1$, it follows

$$2\pi N_G\left(D\right)\le {\displaystyle \frac{T^{ϵ}+O(\Delta L)}{1/2-a}}\ll T^{1/2}{L}^3.$$

i.e.

$${\Delta }_{2T}^{T}argG(1/2+it)\le O\left(T^{1/2}{L}^3\right).$$

and

$$(N\left(2T\right)-N\left(T\right))-(N_0\left(2T\right)-N_0\left(T\right))\le O\left(T^{1/2}{L}^3\right).$$

Then let T be $T/2^k,1\le k\le {log}_2\left(T\right)$, and summing. This proves Theorem 1.1 in the case that there are no zeros of $G\left(s\right)$ on the boundary of D.

For the rest case, let $N_1$ and $N_2$ be the numbers of zeros of $G\left(s\right)$ on the left side of D, $\sigma =1/2$, and in D with $\sigma >1/2$, respectively. Indent the left side of D with small semicircles with centers at the zeros and lying in $\sigma \ge 1/2$. Let $N_1^{\prime }$ be the number of distinct zeros in the $N_1$ zeros. Let $V_j$ be the variation in $argG$ in the jth interval between the successive semicircles. Then by the principle of argument, it has

$$\sum _j{V}_j-\pi N_1=2\pi N_2+O\left(L\right),$$

(3.28)

Let $W_j$ be the variation of argument of

$$h\left(s\right)(f^{\prime }\left(s\right)+f^{\prime }(1-s))G\left(s\right)$$

in the jth interval, where $W_j$ is taken for increasing t, while $V_j$ is taken for decreasing t. With (3.2) and (3.28), it has

$$\begin{array}{cc}\hfill \sum _j{W}_j& =\mathrm{Im}\left(f\right){|}_T^{T+U}-\sum _j{V}_j\hfill \\ \hfill & =\mathrm{Im}\left(f\right){|}_T^{T+U}-(2\pi N_2+\pi N_1)+O\left(L\right)\hfill \end{array}$$

(3.29)

By (3.8), in the jth open interval, the number of zeros of $\zeta (1/2+it)$ is at least

$$(W_j/\pi )-1.$$

and in all the open intervals, the number of zeros is at least

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{\pi }}\sum _j{W}_j-N_1^{\prime }-1& ={\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-(2N_2+N_1)-N_1^{\prime }-1+O\left(L\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-2N_G\left(D\right)+N_1-N_1^{\prime }+O\left(L\right)\hfill \end{array}$$

(3.30)

Moreover, by (3.7), we can know that on the side $\sigma =1/2$, a zero of $G\left(s\right)$ is also a zero of ${\zeta }^{\prime }\left(s\right)$, and so a zero of $\zeta \left(s\right)$, with multiplicity one greater, so there are $N_1+N_1^{\prime }$ such zeros of $\zeta (1/2+it)$, adding to (3.30), in total, it has

$$N_0(T+U)-N_0\left(T\right)\ge {\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-2N_G\left(D\right)+2N_1+O\left(L\right).$$

By (3.11), we can know

$${\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}=N(T+U)-N\left(T\right)+O\left(L\right).$$

i.e.

$$(N(T+U)-N\left(T\right))-(N_0(T+U)-N_0\left(T\right))\le O\left(T^{1/2}{L}^3\right).$$

□

Besides, we know that on the critical line a zero of $G\left(s\right)$ is also a zero of ${\zeta }^{\prime }\left(s\right)$, and so a zero of $\zeta \left(s\right)$, with multiplicity one greater. Hence

$$\sum (m-1)\le N_G\left(D\right).$$

where sum is over the distinct zeros of $\zeta \left(s\right)$ on the left side of D, m is the multiplicity of a zero.

And so,

$$\sum _{m\ge 2}m\le 2N_G\left(D\right)\le O\left(T^{1/2}{L}^3\right).$$

(3.31)

This means that the non-trivial zeros of $\zeta \left(s\right)$ are all on the critical line, and all are simple, with at most $O\left(T^{1/2}{L}^3\right)$ ones excepted.