New Zero-Density Results for Automorphic L-Functions of GL(n)

Abstract

Let $L(s,\pi )$ be an automorphic L-function of $GL\left(n\right)$, where $\pi$ is an automorphic representation of group $GL\left(n\right)$ over rational number field $\mathbb{Q}$. In this paper, we study the zero-density estimates for $L(s,\pi )$. Define $N_{\pi }(\sigma ,T_1,T_2)$ = ♯ {$\rho$ = $\beta$ + $i\gamma$: $L(\rho ,\pi )$ = 0, $\sigma <\beta <1$, $T_1\le \gamma \le T_2$}, where $0\le \sigma <1$ and $T_1<T_2$. We first establish an upper bound for $N_{\pi }(\sigma ,T,2T)$ when $\sigma$ is close to 1. Then we restrict the imaginary part $\gamma$ into a narrow strip $[T,T+T^{\alpha }]$ with $0<\alpha \le 1$ and prove some new zero-density results on $N_{\pi }(\sigma ,T,T+T^{\alpha })$ under specific conditions, which improves previous results when $\sigma$ near $\frac{3}{4}$ and 1, respectively. The proofs rely on the zero detecting method and the Halász-Montgomery method.

1. Introduction

Let $\pi$ be an automorphic representation of group $GL\left(n\right)$ over rational number field $\mathbb{Q}$. Then the automorphic (finite-part) L-function related to $\pi$ can be defined as

$$L(s,\pi )=\sum _{n=1}^{\infty }\frac{A_{\pi }\left(n\right)}{n^s}(\Re s>1).$$

Furthermore, $L(s,\pi )$ satisfies the Euler product

$$L(s,\pi )=\prod _{p<\infty }\prod _{j=1}^m{(1-{\alpha }_{\pi }(p,j)p^{-s})}^{-1}(\Re s>1),$$

(1)

where ${\alpha }_{\pi }(p,j)$ are the Langlands parameters of $\pi$. The automorphic L-function $L(s,\pi )$ can be analytically continued to the whole complex plane and has a standard functional equation.

The zero estimates of $L(s,\pi )$ is an important topic in number theory and has many applications in various problems, for instance, the applications in the composition of integers and integral ideals (see [1,2,3,4], etc.) in divisor problems (see [5,6,7,8], etc.) and in mean value estimates involving Hecke eigenvalues (see [9,10,11,12], etc.). Many scholars have established zero-density estimates of $L(s,\pi )$ (see [13,14,15], etc.). We know that all non-trivial zeros of $L(s,\pi )$ are included in the critical strip $0<\Re s<1$. As we all know, the Generalized Riemann Hypothesis ($GRH$) conjectures that all these non-trivial zeros are on the critical line $\Re s=\frac{1}{2}$. Now the $GRH$ is still open. Then it is natural to study the zero-density estimates of $L(s,\pi )$ in a rectangle including the critical line.

Let

$$N_{\pi }(\sigma ,T_1,T_2)=♯\{\rho =\beta +i\gamma :L(\rho ,\pi )=0,\sigma <\beta <1,T_1\le \gamma \le T_2\},$$

where $0\le \sigma <1$ and $T_1<T_2$. Then the famous “density hypothesis” is

$$N_{\pi }(\sigma ,T,2T)\ll T^{2(1-\sigma )+\epsilon },\frac{1}{2}\le \sigma <1.$$

Here we take the Riemann zeta function $\zeta \left(s\right)$ as an example of the automorphic L-function $L(s,\pi )$. To specify $\pi$ by cusp forms and Maass forms, we refer to the references [16,17,18] and references therein. From the works of Ingham [19] and Huxley [20] we know that

$$\begin{array}{c}\hfill N_{\zeta }(\sigma ,T,2T)\ll \left\{\begin{array}{cc}{T}^{\frac{3(1-\sigma )}{2-\sigma }+\epsilon },\hfill & \mathrm{if}\frac{1}{2}\le \sigma <\frac{3}{4},\hfill \\ T^{\frac{3(1-\sigma )}{3\sigma -1}+\epsilon },\hfill & \mathrm{if}\frac{3}{4}\le \sigma <1.\hfill \end{array}\right.\end{array}$$

About the results on the density hypothesis, in 1977, Jutila [21] proved that

$$N_{\zeta }(\sigma ,T,2T)\ll T^{2(1-\sigma )+\epsilon },\frac{11}{14}\le \sigma <1.$$

In 2000, Bourgain [22] improved Jutila’s result to

$$N_{\zeta }(\sigma ,T,2T)\ll T^{2(1-\sigma )+\epsilon },\frac{25}{32}\le \sigma <1.$$

For $\frac{1}{2}\le u\le 1$, let

$$M(u,T)=\underset{1\le \gamma \le T}{max}\left|\zeta (u+i\gamma )\right|.$$

When $\sigma$ is close to 1, Ivić (see [23], Theorem 11.3) proved that for $\frac{9}{10}\le \sigma \le 1$,

$$N_{\zeta }(\sigma ,-T,T)\ll {\left(M(5\sigma -4,3T)\right)}^{\frac{7}{6}}{log}^{\frac{169}{12}}T.$$

In this paper, motivated by Ivić’s work, we first establish an upper bound of $N_{\pi }(\sigma ,T,2T)$, when $\sigma$ is close to 1 in the following theorem.

Theorem 1.

Let $M_1(u,T)=\underset{1\le \gamma \le T}{max}\left|\zeta (u+i\gamma )\right|$ and $M_2(u,T)=\underset{1\le \gamma \le T}{max}\left|L(u+i\gamma ,\pi )\right|$ for $\frac{1}{2}\le u<1$, $T\ge 3$. Then for $\frac{9}{10}\le \sigma \le 1$ we have

$$N_{\pi }(\sigma ,T,2T)\ll {\left(M_1(5\sigma -4,3T)\right)}^{\frac{2}{3}}{\left(M_2(5\sigma -4,4T)\right)}^{\frac{1}{2}}{T}^{\epsilon }.$$

(2)

Remark 1.

Since $L(s,\pi )$ is a general automorphic L-function in Theorem 1, the mean value estimate now is worse than the case of the Rieman zeta function, which results in the $T^{\epsilon }$ (see the argument around (22) for details).

Now we restrict the imaginary part $\gamma$ into a narrow strip $[T,T+T^{\alpha }]$ and suppose

$${\int }_T^{T+T^{\alpha }}{\left|L\left(\frac{1}{2}+it,\pi \right)\right|}^{2l}dt{\ll }_{\epsilon ,\pi }{T}^{\theta +\epsilon },$$

(3)

for certain $0<\alpha \le 1$ and $\theta \ge \alpha$, where $l\ge 1$ is an integer and $T\ge 3$. Ye and Zhang [24] established some bounds for $N_{\pi }(\sigma ,T,T+T^{\alpha })$ with $\frac{1}{2}\le \sigma <1$. Later, Dong, Liu and Zhang [25] obtained a sharper bound for $N_{\pi }(\sigma ,T,T+T^{\alpha })$ when $\sigma$ is close to 1 and show a range of $\sigma$ for which the density hypothesis holds. We shall keep on studying zero-density estimates for $L(s,\pi )$ in this strip and improve previous results when $\sigma$ near $\frac{3}{4}$ and 1, respectively.

Theorem 2.

Let $L(s,\pi )$ be an automorphic L-function satisfying (3). Then for $\frac{3}{4}\le \sigma \le \frac{10}{13}$ and $2\sigma (2\alpha \theta -16\theta +3{\alpha }^{2}l-6{\alpha }^2)\ge 4\alpha \theta -20\theta +3{\alpha }^{2}l-12{\alpha }^2$ we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{3{\alpha }^2(1-\sigma )}{\alpha (1-\sigma )-5+8\sigma }+\epsilon },$$

and for $\frac{10}{13}<\sigma <1$ and $2\sigma (6\theta \alpha -22\theta +9l{\alpha }^2-18{\alpha }^2)\ge 12\theta \alpha -20\theta +9l{\alpha }^2-36{\alpha }^2$, we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{9{\alpha }^2(1-\sigma )}{3\alpha (1-\sigma )+11\sigma -5}+\epsilon }.$$

Theorem 3.

Let $L(s,\pi )$ be an automorphic L-function satisfying (3). Then for $\frac{3}{4}\le \sigma \le \frac{10}{13}$ and $2\sigma (\alpha \theta -4\theta +2{\alpha }^{2}l-2{\alpha }^2)\ge 2\alpha \theta -5\theta +2{\alpha }^{2}l-4{\alpha }^2$, we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{4{\alpha }^2(1-\sigma )}{2\alpha (1-\sigma )-5+8\sigma }+\epsilon },$$

and for $\frac{10}{13}<\sigma <1$ and $\sigma (6\theta \alpha -12{\alpha }^2-11\theta +12l{\alpha }^2)\ge 6\theta \alpha -12{\alpha }^2-5\theta +6l{\alpha }^2$, we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{12{\alpha }^2(1-\sigma )}{6\alpha (1-\sigma )+11\sigma -5}+\epsilon }.$$

Notation 1.

Throughout this paper, the letter ε represents a sufficiently small positive number, whose value may change from statement to statement. Constants, both explicit and implicit, in Vinogradov symbols ≪ may depend on ε and π.

2. Some Lemmas

Lemma 1.

For $L(s,\pi )$ we have

$$N_{\pi }(0,T,T+1)=\frac{m}{2\pi }logT(1+o\left(1\right)).$$

Proof. 

We can get this lemma by a standard winding number argument on (3) as in Davenport [26] and Rudnick and Sarnak [15]. □

Lemma 2

([27], Lemma 1.7). Let $\xi ,{\phi }_1,\dots ,{\phi }_R$ be arbitrary vectors in an inner-product vector space over $\mathbb{C}$. If $a={\left\{a_n\right\}}_{n=1}^{\infty }$ and $b={\left\{b_n\right\}}_{n=1}^{\infty }$ are two vectors of $\mathbb{C}$, then the inner-product of a and b is defined as

$$(a,b)=\sum _{n=1}^{\infty }{a}_n{\overline{b}}_n.$$

Then we have the inequality

$$\sum _{r\le R}|(\xi ,{\phi }_r)|\le \parallel \xi \parallel {\left(\sum _{r,s\le R}\left|({\phi }_r,{\phi }_s)\right|\right)}^{\frac{1}{2}},$$

(4)

where ${\parallel a\parallel }^2=(a,a).$

Lemma 3

([23], (4.60)). Suppose that $Y,h>0$ and $k\ge 1$ is an integer, we have

$$\sum _{n=1}^{\infty }{e}^{-{\left(\frac{n}{Y}\right)}^h}{d}_k\left(n\right)n^{-s}={\left(2\pi i\right)}^{-1}{\int }_{2-i\infty }^{2+\infty }{\zeta }^k(s+w)Y^w\Gamma (1+\frac{w}{h})w^{-1}dw.$$

(5)

Lemma 4.

For a fixed θ satisfying $\frac{1}{2}<\theta <1$, define

$$S\left(\theta \right)=\sum _{r,s\le R}{\left|\zeta (\theta +i{t}_r-i{t}_s+i{v}^{\prime })\right|}^2,$$

where $T\le t_r\le T+T^{\alpha }$$(1\le r\le R)$ and $v^{\prime }$ is defined by

$$|\zeta (\theta +i{t}_r-i{t}_s+i{v}^{\prime })|=\underset{-{log}^{2}T\le v\le {log}^{2}T}{max}\left|\zeta (\theta +i{t}_r-i{t}_s+iv)\right|.$$

Suppose that $|t_r-t_s|\ge 3{log}^{4}T$ for $r\ne s\le R$, then we have

$$S\left(\frac{3}{4}\right)\ll T^{\epsilon }\left(R^2+R^{\frac{11}{8}}{T}^{\frac{\alpha }{4}}\right).$$

Proof. 

We can get this lemma by following the argument of ([23], Lemma 11.6). The main difference is that the interval of $t_r$ now is $T\le t_r\le T+T^{\alpha }$. □

3. Proof of Theorem 1

We first consider the number of zeros $\rho =\beta +i\gamma$ of $L(s,\pi )$ in the rectangle

$$\sigma <\beta <1,T\le \gamma \le T+T^{\alpha },$$

(6)

where $\sigma \ge \frac{1}{2}$, $T\ge 3$, and $0<\alpha \le 1$. We define ${\mu }_{\pi }\left(n\right)$ by

$$\frac{1}{L(s,\pi )}=\sum _{n=1}^{\infty }\frac{{\mu }_{\pi }\left(n\right)}{n^s}$$

for $\Re s>1$. By the Euler product of $L(s,\pi )$ in (1), we have

$$\frac{1}{L(s,\pi )}=\prod _{p<\infty }\prod _{j=1}^m(1-{\alpha }_{\pi }(p,j)p^{-s}).$$

Consequently ${\mu }_{\pi }\left(n\right)$ is a multiplicative function and

$${\mu }_{\pi }\left(1\right)=1,$$

$${\mu }_{\pi }\left(p^k\right)={(-1)}^k\sum _{1\le j_1<\cdots <j_k\le m}\prod _{v=1}^k{\alpha }_{\pi }(p,j_v)$$

for $k=1,\dots ,m$ and ${\mu }_{\pi }\left(p^k\right)=0$ for $k>m$. Then from $L(s,\pi )·\frac{1}{L(s,\pi )}=1$, we get

$$\sum _{d|n}{\mu }_{\pi }\left(d\right)A_{\pi }\left(\frac{n}{d}\right)=\left\{\begin{array}{cc}1,\hfill & n=1,\hfill \\ 0,\hfill & n>1.\hfill \end{array}\right.$$

Assume further that $s=\sigma +it$, $T\le t\le T+T^{\alpha }$, $1\ll X\ll Y\le T^A$ for some $A>0$, and $X=X\left(T\right)$ and $Y=Y\left(T\right)$ are two parameters to be decided later. Let

$$M_X(s,\pi )=\sum _{n\le X}\frac{{\mu }_{\pi }\left(n\right)}{n^s}.$$

Then we have

$$L(s,\pi )M_X(s,\pi )=\sum _{n=1}^{\infty }\frac{b_{\pi }\left(n\right)}{n^s},$$

(7)

for $\Re s>1$, where

$$b_{\pi }\left(n\right)=\left\{\begin{array}{cccc}& 1,\hfill & \hfill \mathrm{if}& n=1,\hfill \\ & 0,\hfill & \hfill \mathrm{if}& 2\le n\le X,\hfill \\ & \sum _{\genfrac{}{}{0pt}{}{d|n}{d\le X}}{\mu }_{\pi }\left(d\right)A_{\pi }\left(\frac{n}{d}\right),\hfill & \hfill \mathrm{if}& n>X.\hfill \end{array}\right.$$

In terms of (7), for $\sigma >1$ and $\Re s>0$, we obtain

$$\begin{array}{c}\hfill \frac{1}{2\pi i}{\int }_{\sigma }L(s+\omega ,\pi )M_X(s+\omega ,\pi )\Gamma \left(\omega \right)Y^{\omega }d\omega =\frac{1}{2\pi i}{\int }_{\sigma }\sum _{n=1}^{\infty }\frac{b_{\pi }\left(n\right)}{n^{s+\omega }}\Gamma \left(\omega \right)Y^{\omega }d\omega .\end{array}$$

From the inverse Mellin transform of $\Gamma \left(\omega \right)$, the right-hand side of the above formula can be written as

$$\begin{array}{c}\hfill \frac{1}{2\pi i}{\int }_{\sigma }\Gamma \left(\omega \right)Y^{\omega }d\omega +\sum _{n>X}\frac{b_{\pi }\left(n\right)}{n^s}\frac{1}{2\pi i}{\int }_{\sigma }\Gamma \left(\omega \right){\left(\frac{Y}{n}\right)}^{\omega }d\omega =e^{-\frac{1}{Y}}+\sum _{n>X}\frac{b_{\pi }\left(n\right)}{n^s}{e}^{-\frac{n}{Y}}.\end{array}$$

Hence we have, for $\sigma >1$ and $\Re s>0$,

$$e^{-\frac{1}{Y}}+\sum _{n>X}\frac{b_{\pi }\left(n\right)}{n^s}{e}^{-\frac{n}{Y}}=\frac{1}{2\pi i}{\int }_{\sigma }L(s+\omega ,\pi )M_X(s+\omega ,\pi )\Gamma \left(\omega \right)Y^{\omega }d\omega .$$

(8)

Now we move the line of integration of (8) to $\Re \omega =u-\beta <0$ for some suitable $\frac{1}{2}\le u<1$. Then integral on the right hand of (8) becomes

$$L(s,\pi )M_X(s,\pi )+\frac{1}{2\pi i}\underset{(u-\beta )}{\int }L(s+\omega ,\pi )M_X(s+\omega ,\pi )\Gamma \left(\omega \right)Y^{\omega }d\omega ,$$

where $L(s,\pi )M_X(s,\pi )$ is the residue of the integrand at $\omega =0$. Setting $\omega =u-\beta +iv$ and taking s equal to a non-trivial zero $\rho =\beta +i\gamma$, we have

$$\begin{array}{cc}\hfill & e^{-\frac{1}{Y}}+\sum _{n>X}\frac{b_{\pi }\left(n\right)}{n^{\rho }}{e}^{-\frac{n}{Y}}\hfill \\ \hfill & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }L(u+i\gamma +iv,\pi )M_X(u+i\gamma +iv,\pi )\Gamma (u-\beta +iv)Y^{u-\beta +iv}dv.\hfill \end{array}$$

(9)

Since $e^{-\frac{1}{Y}}\to 1$ and ${\displaystyle \sum _{n>Y{log}^{2}Y}}\frac{b_{\pi }\left(n\right)}{n^{\rho }}{e}^{-\frac{n}{Y}}$ contribute $o\left(1\right)$ as $Y\to \infty$, a non-trivial zero $\rho =\beta +i\gamma$ counted in $N_{\pi }(\sigma ,T,T+T^{\alpha })$ satisfies either

$$1\ll \left|\sum _{X<n\le Y{log}^{2}Y}\frac{b_{\pi }\left(n\right)}{n^{\rho }}{e}^{-\frac{n}{Y}}\right|$$

(10)

or

$$1\ll {\int }_{-\infty }^{+\infty }\left|L(u+i\gamma +iv,\pi )M_X(u+i\gamma +iv,\pi )\Gamma (u-\beta +iv)Y^{u-\beta +iv}\right|dv.$$

(11)

For the integral in (11) we have

$${\int }_{-\infty }^{-{log}^{2}T}+{\int }_{{log}^{2}T}^{+\infty }=o\left(1\right)$$

due to the fact that they are absolutely convergent. By Stirling’s formula again, we can further remove the $\Gamma$ function in (11) and get that a non-trivial zero $\rho =\beta +i\gamma$ counted in $N_{\pi }(\sigma ,T,T+T^{\alpha })$ satisfies either (10) or

$$1\ll Y^{u-\beta }{\int }_{-{log}^{2}T}^{{log}^{2}T}\left|L(u+i\gamma +iv,\pi )M_X(u+i\gamma +iv,\pi )\right|dv.$$

(12)

We divide the elongated rectangle (6) into successive rectangles of length $3{log}^{4}T$ noting Lemma 1. These rectangles start at

$$\sigma <\beta <1,T\le \gamma \le T+3{log}^{4}T$$

(13)

and the last one may have a shorter length. Call these smaller rectangles as $I_1,I_2,\dots$. The number of zeros are denoted by $A_{ij}$ in all odd-numbered rectangles if $j=1$ or in all even-numbered rectangles if $j=2$, which satisfy (10) if $i=1$, or (12) if $i=2$. Then, we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\le A_{11}+A_{12}+A_{21}+A_{22}.$$

We can get a sequence of zeros ${\rho }_r^{\left(ij\right)}={\beta }_r^{\left(ij\right)}+i{\gamma }_r^{\left(ij\right)}$ counted in $A_{ij}$ for $r=1,\dots ,R_{ij}$, if we choose a zero from each rectangle which contains at least one zero. Here, $R_{ij}$ is the number of rectangles that contains at least one zero counted in $A_{ij}$. By Lemma 1, we note that each rectangle contains at most $\frac{3m}{2\pi }{log}^{5}T(1+o\left(1\right))$ zeros. Consequently, we have

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\le (R_{11}+R_{12}+R_{21}+R_{22}){log}^{5}T.$$

(14)

Now we begin to estimate $R_{1j}$. By a dyadic subdivision on the sum in (10), we know that each ${\rho }_r^{\left(1j\right)}={\beta }_r^{\left(1j\right)}+i{\gamma }_r^{\left(1j\right)}$ counted in $R_{1j}$ satisfies

$$1\ll \cdots +\left|\sum _{\frac{\sqrt{Y}}{2}<n<\sqrt{Y}}\frac{b_{\pi }\left(n\right)}{n^{{\rho }^{\left(1j\right)}}}{e}^{-\frac{n}{Y}}\right|+\left|\sum _{\sqrt{Y}<n<2\sqrt{Y}}\frac{b_{\pi }\left(n\right)}{n^{{\rho }^{\left(1j\right)}}}{e}^{-\frac{n}{Y}}\right|+\cdots ,$$

(15)

where there are $O(log\left(Y{log}^{2}Y\right))$ terms. Then there exists $M_r=2^{v^r}\sqrt{Y}$ for some $v^r\in \mathbb{Z}$ and $X\le M_r\le Y{log}^{2}Y$ such that

$$\sum _{M_r<n<2M_r}\frac{b_{\pi }\left(n\right)}{n^{{\rho }^{\left(1j\right)}}}{e}^{-\frac{n}{Y}}\gg \frac{1}{logY}.$$

(16)

Raising (16) to kth power we get

$$\sum _{M_r^k<n\le {\left(2M_r\right)}^k}\frac{c_{\pi }\left(n\right)}{n^{{\rho }^{\left(1j\right)}}}\gg \frac{1}{{log}^{k}Y},$$

(17)

with

$$c_{\pi }\left(n\right)=\sum _{\genfrac{}{}{0pt}{}{n=n_1{n}_2\cdots n_k}{M_r<n_i\le 2M_r}}{b}_{\pi }\left(n_1\right)\cdots b_{\pi }\left(n_k\right)e^{-\frac{n_1+n_2+\cdots +n_k}{Y}}$$

and k is a natural number depending on $M_r$ such that $M_r^k=N$, ${\left(2M_r\right)}^k=P\le T^c$, from where $k\ll 1$ and $P\ll N$. We split the sum in (17) into subsums of length N and choose k so that $M_r^k\le Y^r{log}^{2r}Y<M_r^{k+1}$, where r is a fixed integer and $k\ge r\ge 1$ is satisfied. Then (17) can be written as

$$\sum _{N<n\le 2N}\frac{c_{\pi }\left(n\right)}{n^{{\rho }^{\left(1j\right)}}}\gg \frac{1}{{log}^{D}Y}$$

(18)

for some $D\asymp 1$ and

$$Y^{\frac{r^2}{r+1}}{log}^{\frac{2r^2}{r+1}}Y\ll N\ll Y^r{log}^{2r}Y.$$

(19)

By partial summation and (18), we have

$$\begin{array}{cc}\hfill R_{1j}& \ll {log}^{D}T\left\{\sum _{{\rho }^{\left(1j\right)}}\right(\left|\sum _{N<n\le 2N}{c}_{\pi }\left(n\right)n^{-\sigma -i{\gamma }_r^{\left(1j\right)}}\right|N^{\sigma -{\beta }_r^{\left(1j\right)}}\hfill \\ \hfill & +{\int }_N^{2N}\left|\sum _{N<n\le u}{c}_{\pi }\left(n\right)n^{-\sigma -i{\gamma }_r^{\left(1j\right)}}\right|N^{\sigma -{\beta }_r^{\left(1j\right)}-1}du\left)\right\}.\hfill \end{array}$$

(20)

We relabel $c_{\pi }\left(n\right)$ to satisfy $c_{\pi }\left(n\right)=0$ for $n>u$. Then simplifying (20), we can get

$$R_{1j}\ll {log}^{D}T\sum _{r=1}^{R_{1j}}\left|\sum _{N<n\le 2N}{c}_{\pi }\left(n\right)n^{-\sigma -i{\gamma }_r^{\left(1j\right)}}\right|.$$

(21)

From now on, we let $\alpha =1$ in (6). Now we apply Lemma 2 to (21) and take $\xi ={\left\{{\xi }_n\right\}}_{n=1}^{\infty }$ with $c_{\pi }\left(n\right)\ll n^{\epsilon }$, where

$${\xi }_n=\left\{\begin{array}{ccc}& c_{\pi }\left(n\right){(e^{-\frac{n}{2N}}-e^{-\frac{n}{N}})}^{-\frac{1}{2}}{n}^{-\sigma },\hfill & \hfill N<n\le 2N,\\ & 0,\hfill & \hfill otherwise,\end{array}\right.$$

and ${\phi }_r={\left\{{\phi }_{r,n}\right\}}_{n=1}^{\infty }$ with ${\phi }_{r,n}={(e^{-\frac{n}{2N}}-e^{-\frac{n}{N}})}^{\frac{1}{2}}{n}^{-i{\gamma }_r^{\left(1j\right)}}(n=1,2,3\dots )$. Denoting the imaginary part of representative zeros of $R_{1j}$ by ${\gamma }_1^{\left(1j\right)},\dots ,{\gamma }_{R_{1j}}^{\left(1j\right)}$, we then have

$$\begin{array}{cc}\hfill R_{1j}^2& \ll {log}^{2D}T\sum _{r\le R_{1j}}|\sum _{N<n\le 2N}{\xi }_n{\phi }_{r,n}{|}^2\ll {log}^{2D}{T\parallel \xi \parallel }^2\sum _{r,s\le R_{1j}}\left|({\phi }_r,{\phi }_s)\right|\hfill \\ \hfill & \ll {log}^{2D}T\left(\sum _{N<n\le 2N}{c}_{\pi }^2\left(n\right)e^{-\frac{2n}{Y}}{n}^{-2\sigma }\right)\left(\sum _{r=s}|({\phi }_r,{\phi }_s)|+\sum _{r\ne s}\left|({\phi }_r,{\phi }_s)\right|\right)\hfill \\ \hfill & \ll {log}^{2D}T\left(\sum _{N<n\le 2N}{c}_{\pi }^2\left(n\right)n^{-2\sigma }{e}^{-\frac{2n}{Y}}\right)\left(R_{1j}N+\sum _{r\ne s\le R_{1j}}\left(H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right)\right)\hfill \\ \hfill & \ll T^{\epsilon }\left(N^{2-2\sigma }{R}_{1j}+\sum _{r\ne s\le R_{1j}}\left(H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right)\right),\hfill \end{array}$$

(22)

where

$$\begin{array}{cc}\hfill H\left(it\right)& =\sum _{n=1}^{\infty }(e^{-\frac{n}{2N}}-e^{-\frac{n}{N}})n^{-it}\hfill \\ \hfill & =\frac{1}{2\pi i}{\int }_{2-i\infty }^{2+i\infty }\zeta (w+it)({\left(2N\right)}^w-N^w)\Gamma \left(w\right)dw,\hfill \end{array}$$

(23)

since $1\ll e^{-\frac{n}{2N}}-e^{-\frac{n}{N}}\ll 1$ for $N<n\le 2N$, ${\parallel \xi \parallel }^2\ll T^{\epsilon }{N}^{1-2\sigma }$ and $H\left(0\right)\ll N$. Thus, we have

$$H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})=\frac{1}{2\pi i}{\int }_{2-i\infty }^{2+i\infty }\zeta (w+i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\left({\left(2N\right)}^w-N^w\right)\Gamma \left(w\right)dw.$$

(24)

Moving the line of integration in (24) to $\Re w=u<1$, we encounter a simple pole at $w=1-i{\gamma }_r^{\left(1j\right)}+i{\gamma }_s^{\left(1j\right)}$ with residue $\ll N{e}^{-|{\gamma }_s^{\left(1j\right)}-{\gamma }_r^{\left(1j\right)}|}$, so that

$$\begin{array}{cc}\hfill & H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\hfill \\ \hfill & =\frac{1}{2\pi i}{\int }_{u-i\infty }^{u+i\infty }\zeta (w+i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\left({\left(2N\right)}^w-N^w\right)\Gamma \left(w\right)dw+O\left(N{e}^{-|{\gamma }_s^{\left(1j\right)}-{\gamma }_r^{\left(1j\right)}|}\right).\hfill \end{array}$$

(25)

In view of $|\Gamma \left(s\right)|={\left(2\pi \right)}^{\frac{1}{2}}{\left|t\right|}^{\sigma -\frac{1}{2}}{e}^{-\frac{\pi t}{2}}{(1+O(\left|t\right|}^{-1}\left)\right)$, the integral in (25) is $o\left(1\right)$ for $N\ll T^c$ if $|\Im w|\ge {log}^{2}T$, which gives

$$\begin{array}{cc}\hfill \sum _{r\ne s\le R_{1j}}\left|H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right|& \ll N\sum _{r\ne s\le R_{1j}}{e}^{-|{\gamma }_s^{\left(1j\right)}-{\gamma }_r^{\left(1j\right)}|}+o\left(R_{1j}^2\right)\hfill \\ \hfill & +N^u{\int }_{-{log}^{2}T}^{{log}^{2}T}\sum _{r\ne s\le R_{1j}}\left|\zeta (u+i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)}+iv)\right|dv.\hfill \end{array}$$

(26)

Since the ${{\gamma }_r^{\left(1j\right)}}^{\prime }s$ are at least $3{log}^{4}T$ apart, the first sum on the right side of (26) is $O\left(R_{1j}\right)$. For the second sum on the right side of (26), we fix each s and let ${\tau }_r={\gamma }_r^{\left(1j\right)}-{\gamma }_s^{\left(1j\right)}+v$. Then we have $|{\tau }_r|⩽3T$ for $r=1,2,\dots ,R_{1j}$ and $|{\tau }_{r_1}-{\tau }_{r_2}|\ge 3{log}^{4}T$ for $r_1\ne r_2$. Hence we get

$$\sum _{r\ne s\le R_{1j}}\left(H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right)\ll R_{1j}^2{N}^u{M}_1(u,3T){log}^{2}T.$$

(27)

Inserting (27) into (22) we obtain

$$\begin{array}{c}\hfill R_{1j}^2\ll T^{\epsilon }\left(R_{1j}{N}^{2-2\sigma }+R_{1j}^2{N}^{1+u-2\sigma }{M}_1(u,3T)\right).\end{array}$$

Then we have for $\sigma \ge \frac{u+1}{2}$,

$$R_{1j}\ll \underset{X\le N\le Y{log}^{2}Y}{max}{N}^{2-2\sigma }{T}^{\epsilon }\ll Y^{2-2\sigma }{T}^{\epsilon },$$

(28)

if $X^{2\sigma -1-u}\gg M_1(u,3T)T^{\epsilon }$.

Now we turn to estimate $R_{2j}$. We may suppose first that $\sigma <1-\frac{C}{logT}$ in view of the zero-free region. Note that in this case the zeros ${\rho }_r^{\left(2j\right)}={\beta }_r^{\left(2j\right)}+i{\gamma }_r^{\left(2j\right)}$, $j=1,2$ satisfy

$$1\ll Y^{u-\beta }{\int }_{-{log}^{2}T}^{{log}^{2}T}\left|L(u+i{\gamma }_r^{\left(2j\right)}+iv,\pi )M_X(u+i{\gamma }_r^{\left(2j\right)}+iv,\pi )\right|dv.$$

(29)

From the mean value theorem for integration, we can see that there exists $t_r^{\left(2j\right)}={\gamma }_r^{\left(2j\right)}+v$, which makes (29) become

$$1\ll Y^{u-\beta }\left|L(u+i{t}_r^{\left(2j\right)},\pi )M_X(u+i{t}_r^{\left(2j\right)},\pi )\right|{log}^{2}T.$$

(30)

Thus, from (30) we get

$$M_X(u+i{t}_r^{\left(2j\right)},\pi )\gg Y^{\sigma -u}{\left(M_2(u,4T){log}^{2}T\right)}^{-1}$$

for $R_{2j}$ points $t_r^{\left(2j\right)}$ such that $|t_r^{\left(2j\right)}|⩽4T$. Then there is a number N$(1\ll N\le X)$ such that

$$\sum _{N<n\le 2N}\frac{{\mu }_{\pi }\left(n\right)}{n^{u+i{t}_r^{\left(2j\right)}}}\gg Y^{\sigma -u}{\left(M_2(u,4T){log}^{2}T\right)}^{-1},$$

hence it is easy to get

$$1\ll Y^{u-\sigma }{M}_2(u,4T){log}^{2}T\sum _{N<n\le 2N}\frac{{\mu }_{\pi }\left(n\right)}{n^{u+i{t}_r^{\left(2j\right)}}}.$$

We can apply Lemma 2 as in the previous case, but now the choice of $\xi$ and ${\phi }_r$ are different. We shall take $\xi ={\left\{{\xi }_n\right\}}_{n=1}^{\infty }$ with ${\xi }_n={\mu }_{\pi }\left(n\right){(e^{-\frac{n}{2N}}-e^{-\frac{n}{N}})}^{-\frac{1}{2}}{n}^{-u}$ for $N<n\le 2N$ and zero otherwise, ${\phi }_r={\left\{{\phi }_{r,n}\right\}}_{n=1}^{\infty }$ with ${\phi }_{r,n}={(e^{-\frac{n}{2N}}-e^{-\frac{n}{N}})}^{\frac{1}{2}}{n}^{-i{t}_r^{\left(2j\right)}}(n=1,2,3\dots )$. Then

$$\begin{array}{cc}\hfill R_{2j}^2\ll & Y^{2u-2\sigma }{M_2}^2(u,4T){log}^{4}T{\left(\sum _{r\le R_{2j}}\left|\sum _{N<n\le 2N}\frac{{\mu }_{\pi }\left(n\right)}{n^{u+i{t}_r^{\left(2j\right)}}}\right|\right)}^2\hfill \\ \hfill \ll & Y^{2u-2\sigma }{M_2}^2(u,4T){log}^4{T\parallel \xi \parallel }^2\sum _{r,s\le R_{2j}}\left|({\phi }_r,{\phi }_s)\right|\hfill \\ \hfill \ll & Y^{2u-2\sigma }{M_2}^2(u,4T)N^{1-2u+\epsilon }{log}^{4}T\left(R_{2j}N+\sum _{r\ne s\le R_{2j}}H(i{t}_r^{\left(2j\right)}-i{t}_s^{\left(2j\right)})\right)\hfill \\ \hfill \ll & Y^{2u-2\sigma }{M_2}^2(u,4T)N^{2-2u}{R}_{2j}{log}^{5}T\hfill \\ \hfill +& Y^{2u-2\sigma }{M_2}^2(u,4T)M_1(u,3T)N^{1-u}{R}_{2j}^2{log}^{7}T,\hfill \end{array}$$

Thus, we have

$$R_{2j}\ll X^{2-2u}{Y}^{2u-2\sigma }{M_2}^2(u,4T){log}^{5}T$$

if

$$Y^{2\sigma -2u}\gg M_1(u,3T){M_2}^2(u,4T)X^{1-u}{log}^{7}T.$$

Therefore X and Y can be chosen as

$$X=C_1{M}_1{(u,3T)}^{\frac{1}{2\sigma -1-u}}{T}^{\epsilon },$$

$$Y=C_2{M_1(u,3T)}^{\frac{1}{2\sigma -u-1}}{M_2(u,4T)}^{\frac{1}{\sigma -u}}{T}^{\epsilon }.$$

We see that the bound for $R_{2j}$ is smaller than the one for $R_{1j}$. Recalling (14) we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,2T)& \ll Y^{2-2\sigma }{T}^{\epsilon }\hfill \\ \hfill & \ll {\left({M_1(u,3T)}^{\frac{1}{2\sigma -u-1}}{M_2(u,4T)}^{\frac{1}{\sigma -u}}\right)}^{2-2\sigma }{T}^{\epsilon },\hfill \end{array}$$

(31)

where ${\displaystyle \frac{1}{2}}⩽u⩽1$, $\frac{u+1}{2}⩽\sigma <1-\frac{C}{logT}$.

Let $u=k\sigma -(k-1)$$\left(k>2\right)$. Then in view of (31), we see that it is appropriate to take $k=5$. Hence $u=5\sigma -4$, $\sigma \ge \frac{u+1}{2}$ is satisfied and $u\ge \frac{1}{2}$ for $\sigma \ge \frac{9}{10}$. Thus, Theorem 1 follows from (31) with $u=5\sigma -4$.

4. Proof of Theorem 2

Throughout the proof of Theorem 2, we restrict the range of zeros to (6). From the estimates of $R_{1j}^2$ in (22), we obtain now

$$R_{1j}^2\ll T^{\epsilon }\left(N^{2-2\sigma }{R}_{1j}+N^{1-2\sigma }\sum _{r\ne s\le R_{1j}}\left|H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right|\right).$$

(32)

Recall the functional equation $\zeta \left(s\right)=\chi \left(s\right)\zeta (1-s)$ with

$$\chi \left(s\right)={\left(\frac{2\pi }{t}\right)}^{\sigma +it-\frac{1}{2}}{e}^{i(t+\frac{\pi }{4})}(1+O|t^{-1}\left|\right).$$

Moving the line of integration in $H\left(it\right)$ to $\Re w=\frac{1}{4}$ and applying the Cauchy–Schwarz inequality, we get

$$\begin{array}{cc}\hfill \sum _{r\ne s\le R_{1j}}\left|H(i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)})\right|\ll & N\sum _{r\ne s\le R_{1j}}{e}^{-|{\gamma }_r^{\left(1j\right)}-{\gamma }_s^{\left(1j\right)}|}+R_{1j}^2+N^{\frac{1}{4}}{T}^{\frac{\alpha }{4}}\hfill \\ \hfill \times & {\int }_{-{log}^{2}T}^{{log}^{2}T}\sum _{r\ne s\le R_{1j}}\left|\zeta (\frac{3}{4}+i{\gamma }_r^{\left(1j\right)}-i{\gamma }_s^{\left(1j\right)}+iv)\right|dv\hfill \\ \hfill \ll & R_{1j}N+R_{1j}^2+T^{\epsilon +\frac{\alpha }{4}}{N}^{\frac{1}{4}}{R}_{1j}{\left(S\left(\frac{3}{4}\right)\right)}^{\frac{1}{2}}\hfill \end{array}$$

(33)

where the $S\left(\frac{3}{4}\right)$ is from Lemma 4.

Substituting (33) into (32), we can get

$$R_{1j}\ll T^{\epsilon }{N}^{2-2\sigma }+R_{1j}{T}^{\frac{\alpha }{4}+\epsilon }{N}^{\frac{5-8\sigma }{4}}+T^{\frac{6\alpha }{5}+\epsilon }{N}^{\frac{20-32\sigma }{5}},$$

(34)

where we used Lemma 4 to $S\left(\frac{3}{4}\right)$. For $R_0$ points lying in an interval of length $T=T_0=N^{\frac{8\sigma -5}{\alpha }-\epsilon }$ we have

$$R_0\ll T^{\epsilon }(N^{2-2\sigma }+T_0^{\frac{6\alpha }{5}}{N}^{\frac{20-32\sigma }{5}})\ll T^{\epsilon }{N}^{2-2\sigma }for\frac{3}{4}\le \sigma \le \frac{10}{13}.$$

In (19), we take $r=2$ to get $Y^{\frac{4}{3}}{log}^{\frac{8}{3}}Y\ll N\ll Y^2{log}^{4}Y$ and then

$$R_{1j}\ll R_0\left(1+\frac{T^{\alpha }}{T_0}\right)\ll T^{\epsilon }{N}^{2-2\sigma }\left(1+\frac{T^{\alpha }}{T_0}\right)\ll T^{\epsilon }\left(Y^{4-4\sigma }+T^{\alpha }{Y}^{\frac{8\alpha (1-\sigma )+20-32\sigma }{3\alpha }}\right)$$

It follows from ([24], (5.21)) that

$$R_{2j}\ll T^{\theta +\epsilon }{Y}^{l(1-2\sigma )}.$$

(35)

Consequently, we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }\sum _{j=1}^2(R_{1j}+R_{2j})\hfill \\ \hfill & \ll T^{\epsilon }(T^{\theta }{Y}^{l(1-2\sigma )}+Y^{4-4\sigma }+T^{\alpha }{Y}^{\frac{8\alpha (1-\sigma )+20-32\sigma }{3\alpha }}).\hfill \end{array}$$

(36)

We set $Y^{4-4\sigma }=T^{\alpha }{Y}^{\frac{8\alpha (1-\sigma )+20-32\sigma }{3\alpha }}$, which is equivalent to $Y=T^{\frac{3{\alpha }^2}{4\alpha (1-\sigma )-20+32\sigma }}$. Thus, from (36) we get

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }(Y^{4-4\sigma }+T^{\theta }{Y}^{l(1-2\sigma )})\hfill \\ \hfill & =T^{\epsilon }(T^{\frac{3{\alpha }^2(1-\sigma )}{\alpha (1-\sigma )-5+8\sigma }}+T^{\theta +\frac{3{\alpha }^{2}l(1-2\sigma )}{4\alpha (1-\sigma )-20+32\sigma }}).\hfill \end{array}$$

(37)

Comparing the two terms in (37), we can get

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{3{\alpha }^2(1-\sigma )}{\alpha (1-\sigma )-5+8\sigma }+\epsilon }$$

for $2\sigma (2\alpha \theta -16\theta +3{\alpha }^{2}l-6{\alpha }^2)\ge 4\alpha \theta -20\theta +3{\alpha }^{2}l-12{\alpha }^2$, from which we complete the proof of the first result of Theorem 2.

For $R_0$ points lying in an interval of length $T=T_0=N^{\frac{11\sigma -5}{3\alpha }}$, we have

$$R_0\ll T^{\epsilon }(N^{2-2\sigma }+R_0{T}_0^{\frac{\alpha }{4}}{N}^{\frac{5-8\sigma }{4}})=T^{\epsilon }(N^{2-2\sigma }+R_0{N}^{\frac{10-13\sigma }{12}}),$$

which implies

$$R_0\ll T^{\epsilon }\frac{N^{2-2\sigma }}{1-N^{\frac{10-13\sigma }{12}}}\ll T^{\epsilon }{N}^{2-2\sigma }\mathrm{for}\frac{10}{13}<\sigma <1.$$

Then we have

$$\begin{array}{cc}\hfill R_{1j}& \ll R_0(1+\frac{T^{\alpha }}{T_0})\ll T^{\epsilon }(N^{2-2\sigma }+T^{\alpha }{N}^{\frac{6\alpha (1-\sigma )-11\sigma +5}{3\alpha }})\hfill \\ \hfill & \ll T^{\epsilon }(Y^{4-4\sigma }+T^{\alpha }{Y}^{\frac{24\alpha (1-\sigma )-44\sigma +20}{9\alpha }}).\hfill \end{array}$$

(38)

Consequently, recalling (35) we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll \sum _{j=1}^2(R_{1j}+R_{2j}){log}^{5}T\hfill \\ \hfill & \ll T^{\epsilon }(Y^{4-4\sigma }+T^{\alpha }{Y}^{\frac{24\alpha (1-\sigma )-44\sigma +20}{9\alpha }}+T^{\theta }{Y}^{l(1-2\sigma )}).\hfill \end{array}$$

(39)

We choose $Y^{4-4\sigma }=T^{\alpha }{Y}^{\frac{24\alpha (1-\sigma )-44\sigma +20}{9\alpha }}$, which is equivalent to $Y=T^{\frac{9{\alpha }^2}{12\alpha (1-\sigma )+4(11\sigma -5)}}$. Thus, from (39) we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }(Y^{4-4\sigma }+T^{\theta }{Y}^{l(1-2\sigma )})\hfill \\ \hfill & =T^{\epsilon }(T^{\frac{9{\alpha }^2(1-\sigma )}{3\alpha (1-\sigma )+11\sigma -5}}+T^{\theta +\frac{9{\alpha }^{2}l(1-2\sigma )}{12\alpha (1-\sigma )+4(11\sigma -5)}}).\hfill \end{array}$$

(40)

Comparing the two terms in (40), we can get

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{9{\alpha }^2(1-\sigma )}{3\alpha (1-\sigma )+11\sigma -5}+\epsilon }$$

for $2\sigma (18{\alpha }^2-6\theta \alpha +22\theta -9l{\alpha }^2)\le 36{\alpha }^2-12\theta \alpha +20\theta -9l{\alpha }^2$, from which we complete the proof of the second result of Theorem 2.

5. Proof of Theorem 3

The proofs of Theorems 2 and 3 are similar, and the main difference is the range of N. Here for completeness, we state some critical details. We let $r=1$ in (19) and get $Y^{\frac{1}{2}}logY\ll N\ll Y{log}^{2}Y$.

Unlike the previous (34), we now have

$$R_{1j}\ll T^{\epsilon }{N}^{2-2\sigma }+R_{1j}{T}^{\frac{\alpha }{4}+\epsilon }{N}^{\frac{5-8\sigma }{4}}+T^{\frac{6\alpha }{5}+\epsilon }{N}^{\frac{20-32\sigma }{5}}.$$

For $R_0$ points lying in an interval of length $T=T_0=N^{\frac{8\sigma -5}{\alpha }-\epsilon }$, we have

$$R_0\ll T^{\epsilon }(N^{2-2\sigma }+T_0^{\frac{6\alpha }{5}}{N}^{\frac{20-32\sigma }{5}})\ll T^{\epsilon }{N}^{2-2\sigma }for\frac{3}{4}\le \sigma \le \frac{10}{13}.$$

Then

$$R_{1j}\ll R_0(1+\frac{T^{\alpha }}{T_0})\ll T^{\epsilon }{N}^{2-2\sigma }(1+\frac{T^{\alpha }}{T_0})\ll T^{\epsilon }(Y^{2-2\sigma }+T^{\alpha }{Y}^{\frac{2\alpha (1-\sigma )+5-8\sigma }{2\alpha }})$$

Consequently, recalling (35), we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }\sum _{j=1}^2(R_{1j}+R_{2j})\hfill \\ \hfill & \ll T^{\epsilon }(T^{\theta }{Y}^{l(1-2\sigma )}+Y^{2-2\sigma }+T^{\alpha }{Y}^{\frac{2\alpha (1-\sigma )+5-8\sigma }{2\alpha }}).\hfill \end{array}$$

(41)

We set $Y^{2-2\sigma }=T^{\alpha }{Y}^{\frac{2\alpha (1-\sigma )+5-8\sigma }{2\alpha }}$ which is equivalent to $Y=T^{\frac{2{\alpha }^2}{2\alpha (1-\sigma )-5+8\sigma }}$. Thus, (41) becomes

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }(Y^{2-2\sigma }+T^{\theta }{Y}^{l(1-2\sigma )})\hfill \\ \hfill & =T^{\epsilon }(T^{\frac{4{\alpha }^2(1-\sigma )}{2\alpha (1-\sigma )-5+8\sigma }}+T^{\theta +\frac{2{\alpha }^{2}l(1-2\sigma )}{2\alpha (1-\sigma )-5+8\sigma }}).\hfill \end{array}$$

(42)

Comparing the two terms in (42) we can get

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{4{\alpha }^2(1-\sigma )}{2\alpha (1-\sigma )-5+8\sigma }+\epsilon }$$

for $2\sigma (\alpha \theta -4\theta +2{\alpha }^{2}l-2{\alpha }^2)\ge 2\alpha \theta -5\theta +2{\alpha }^{2}l-4{\alpha }^2$, from which we complete the proof of the first result of Theorem 3.

For $R_0$ points lying in an interval of length $T=T_0=N^{\frac{11\sigma -5}{3\alpha }}$ we have

$$R_0\ll T^{\epsilon }(N^{2-2\sigma }+R_0{T}_0^{\frac{\alpha }{4}}{N}^{\frac{5-8\sigma }{4}})=T^{\epsilon }(N^{2-2\sigma }+R_0{N}^{\frac{10-13\sigma }{12}}),$$

which implies

$$R_0\ll T^{\epsilon }\frac{N^{2-2\sigma }}{1-N^{\frac{10-13\sigma }{12}}}\ll T^{\epsilon }{N}^{2-2\sigma }for\frac{10}{13}<\sigma \le 1.$$

Then we have

$$\begin{array}{cc}\hfill R_{1j}& \ll R_0(1+\frac{T^{\alpha }}{T_0})\ll T^{\epsilon }(N^{2-2\sigma }+T^{\alpha }{N}^{\frac{6\alpha (1-\sigma )-11\sigma +5}{3\alpha }})\hfill \\ \hfill & \ll T^{\epsilon }(Y^{2-2\sigma }+T^{\alpha }{Y}^{\frac{6\alpha (1-\sigma )-11\sigma +5}{6\alpha }}).\hfill \end{array}$$

(43)

Consequently, recalling (35) we have

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll \sum _{j=1}^2(R_{1j}+R_{2j})\hfill \\ \hfill & \ll T^{\epsilon }(Y^{2-2\sigma }+T^{\alpha }{Y}^{\frac{6\alpha (1-\sigma )-11\sigma +5}{6\alpha }}+T^{\theta }{Y}^{l(1-2\sigma )}).\hfill \end{array}$$

(44)

We choose $Y^{2-2\sigma }=T^{\alpha }{Y}^{\frac{6\alpha (1-\sigma )-11\sigma +5}{6\alpha }}$ which is equivalent to $Y=T^{\frac{6{\alpha }^2}{6\alpha (1-\sigma )+11\sigma -5}}$. Thus, (44) becomes

$$\begin{array}{cc}\hfill N_{\pi }(\sigma ,T,T+T^{\alpha })& \ll T^{\epsilon }(Y^{2-2\sigma }+T^{\theta }{Y}^{l(1-2\sigma )})\hfill \\ \hfill & =T^{\epsilon }(T^{\frac{12{\alpha }^2(1-\sigma )}{6\alpha (1-\sigma )+11\sigma -5}}+T^{\theta +\frac{6{\alpha }^{2}l(1-2\sigma )}{6\alpha (1-\sigma )+11\sigma -5}}.\hfill \end{array}$$

(45)

Comparing the two terms in (45) we can get

$$N_{\pi }(\sigma ,T,T+T^{\alpha })\ll T^{\frac{12{\alpha }^2(1-\sigma )}{6\alpha (1-\sigma )+11\sigma -5}+\epsilon },$$

for $\sigma (6\theta \alpha -12{\alpha }^2-11\theta +12{\alpha }^{2}l)\ge 6\theta \alpha -12{\alpha }^2-5\theta +6{\alpha }^{2}l$, from which we complete the proof of the second result of Theorem 3.