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\begin{document}

\keywords{Lindel\"{o}f conjecture, Riemann zeta function, nontrivial zeros, Riemann theorem, moment conjecture}

\subjclass[2020]{11M06, 11M26}

\title[Lindel\"{o}f conjecture]{A direct approach for the Lindel\"{o}f conjecture related to theory
of the Riemann zeta function}

\author{Xiao-Jun Yang$^{1,2,3}$}

\email{dyangxiaojun@163.com; xjyang@cumt.edu.cn}

\address{$^{1}$ School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China}
\address{$^{2}$ State Key Laboratory for Geomechanics and Deep Underground Engineering,
China University of Mining and Technology, Xuzhou 221116, China}
\address{$^{3}$ School of Mechanics and Civil Engineering, China University of Mining and Technology, Xuzhou 221116, China}


\begin{abstract}
It is due to Littlewood that the
truth of the Riemann theorem implies that of the Lindel\"{o}f conjecture.
This paper aims to use the idea of Littlewood to prove the Lindel\"{o}f
conjecture for the Riemann zeta function.
The Lindel\"{o}f $\mu $ function at the critical line is
zero, with use of the Riemann theorem for the entire Riemann zeta function,
proved based on the work of Heath-Brown. Our result is given to show that the
Lindel\"{o}f conjecture, connected with the proof of
the moment conjecture, is true.
\end{abstract}



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\section{Introduction} {\label{sec:1}}
The Lindel\"{o}f conjecture via Riemann zeta function,
proposed in 1908 by Finnish mathematician Ernst Leonard Lindel\"{o}f, has been
one of most important open problems in the history of mathematics~\cite{1}. More
important, the Lindel\"{o}f conjecture is not only linked with the
consequence of the Riemann conjecture~\cite{2} but also used to investigate the
higher movement for the Riemann zeta function~\cite{3}. The Lindel\"{o}f
conjecture has played the important role in the field of the analytic number
theory~\cite{4}.

Suppose that $\mathcal{C}$, $\mathcal{R}$ and $\mathcal{N}$ are the sets of the complex
numbers, real numbers and natural numbers, respectively. Let $s=\sigma
\mbox{+}it\in \mathcal{C}$ such that $Re\left( s \right)=\sigma \in \mathcal{R}$ and
$Im\left( s \right)=t\in \mathcal{R}$ are the real and imaginary parts of the
complex variable $s$, where $i=\sqrt {-1} $. Let As is well known, the
Riemann zeta function $\zeta \left( s \right)$ of the complex variable
$s=\sigma \mbox{+}it$ is defined by the sum~\cite{5}
\begin{equation}
\label{eq1}
\zeta \left( s \right)=\sum\limits_{k=1}^\infty {k^{-s}} ,
\end{equation}
where $k\in \mathcal{N}$ and $Re\left( s \right)>1$. As is stated in~\cite{5} that
this allows Eq.~(\ref{eq1}) to be a meromorphic continuation to the entire complex
plane $s$, with pole of residue $1$ at $s=1$. The trivial zeros for Eq.~(\ref{eq1})
reads $s=-2\nu $ with $\nu \in \mathcal{N}$. The nontrivial zeros for Eq. (\ref{eq1})
are located on the critical line $Re\left( s \right)=1/2$ and in the
critical trip $0<Re\left( s \right)<1$~\cite{6,7}. The entire Riemann zeta
function $\xi \left( s \right)$ is expressed by the product of the Riemann
zeta function $\zeta \left( s \right)$ or the series~\cite{5,8}:
\begin{equation}
\label{eq2}
\xi \left( s \right)=\zeta \left( s \right)\Pi \left( s
\right)=\sum\limits_{h=1}^\infty {\vartheta \left( h \right)\left(
{s-\frac{1}{2}} \right)^{2h}} ,
\end{equation}
where $\Gamma $ is the gamma function~\cite{9},
\begin{equation}
\label{eq3}
\Pi \left( s \right)=\left( {s-1} \right)\pi ^{-s/2}\Gamma \left( {s/2+1}
\right),
\end{equation}
and~\cite{8}
\begin{equation}
\label{eq4}
\vartheta \left( h \right)=\frac{4}{\left( {2h}
\right)!}\int\limits_1^\infty {\frac{d\left( {z^{\frac{3}{2}}\psi ^{\left( 1
\right)}\left( x \right)} \right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log
x}{2}} \right)^{2h}dx}
\end{equation}
with $h\in \mathcal{R}\cup \left\{ 0 \right\}$ and $\psi \left( x
\right)=\sum\limits_{\upsilon =1}^\infty {e^{-\upsilon ^2\pi x}} $.

Based on the above-mentioned results, it is stated in 2005 by Heath-Brown
~\cite{7} that an equivalent statement for the Riemann theorem~\cite{5} is given as
follows:
\begin{theorem}
\label{T1}
\textbf{Riemann theorem (Heath-Brown statement)}
The Heath-Brown statem states the real part of all zeros of $\xi \left( s
\right)$ is $1/2$.
\end{theorem}

It is equivalent to the Riemann statement~\cite{5} that the real part of the
nontrivial zeros of $\zeta \left( s \right)$ is $1/2$. It is known that Eq.~(\ref{eq2})
is the entire function of order $1$~\cite{10}. Although the Riemann statement
has been achieved in~\cite{11}, we would like to give the proof of the
Heath-Brown statement for the Riemann theorem. For more details for the
zeros, number of zeros and imaginary parts of zeros for $\xi \left( s
\right)$, see~\cite{8,10}.

Based on the above, the Lindel\"{o}f conjecture~\cite{1} claims that for every
positive $\varepsilon >0$,
\begin{equation}
\label{eq5}
\zeta \left( {\sigma \mbox{+}it} \right)\ll t^\varepsilon ,
\end{equation}
where $t\to \infty $.

This easily yields that the equivalent statement for the Lindel\"{o}f
conjecture as follows~\cite{8,10,12}:
\begin{conjecture}
\label{C1}
\textbf{Lindel\"{o}f conjecture }

There exists
\begin{equation}
\label{eq6}
\zeta \left( {\frac{1}{2}\mbox{+}it} \right)\ll t^\varepsilon
\end{equation}
for every positive $\varepsilon >0$, and $t\to \infty $.
\end{conjecture}

There are a number of the equivalent statements for the Lindel\"{o}f
conjecture. The equivalences of these various assertions were proposed in
1915 by Hardy and Riesz~\cite{12} and reported in 2015 by Conrey~\cite{13}. It is
shown in 1912 by Littlewood that \textit{Conjecture \ref{C1}} is the
consequences of \textit{Theorem \ref{T1}} implies~\cite{2}. In 1923, Hardy and
Littlewood give two equivalences for \textit{Conjecture \ref{C1}} states that
~\cite{3}
\begin{equation}
\label{eq7}
\frac{1}{T}\int\limits_1^T {\left| {\zeta \left( {\frac{1}{2}\mbox{+}it}
\right)} \right|^{2m}} dt\ll T^\varepsilon
\end{equation}
for $\varepsilon >0$, $m\in \mathcal{N}$ and $T\to \infty $, and
\begin{equation}
\label{eq8}
\frac{1}{T}\int\limits_1^T {\left| {\zeta \left( {\sigma \mbox{+}it}
\right)} \right|^{2m}} dt\ll T^\varepsilon
\end{equation}
for $\varepsilon >0$, $m\in \mathcal{N}$, $\sigma \ge \frac{1}{2}$ and $T\to
\infty $.

It is stated in 2006 by Laurincikas and Steuding that the equivalence for
the Lindel\"{o}f conjecture becomes~\cite{14}
\begin{equation}
\label{eq9}
\int\limits_0^T {\left| {\zeta \left( {\frac{1}{2}\mbox{+}it} \right)}
\right|^{2m}} dt\ll T^{\varepsilon +1},
\end{equation}
for $\varepsilon >0$, $m\in \mathcal{N}$ and $T\to \infty $.

Eq.~(\ref{eq9}) implies that the moment conjecture states~\cite{15}
\begin{equation}
\label{eq10}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\frac{1}{2}\mbox{+}it}
\right)} \right|^{2m}} dt\ll T^\varepsilon ,
\end{equation}
for $\varepsilon >0$, $m\in \mathcal{N}$ and $T\to \infty $.

There exist the advances for the Lindel\"{o}f conjecture, reported in 2006
by Conrey and Ghosh~\cite{16} and made in 2019 by Fokas~\cite{17} based on the
estimation of relevant exponential sums.

Note that $\mu \left( \sigma \right)$ is the Lindel\"{o}f $\mu $ function,
expressed in Eq.~(\ref{eq5})~\cite{1}. It was proved that
(see~\cite{2,18};~\cite{12},p.18)
\begin{equation}
\label{eq11}
\mu \left( \sigma \right)=\left\{ {\begin{array}{l}
 0,\mbox{ for }\sigma \ge \mbox{1/2,} \\
 1/2-\sigma ,\mbox{ for }\sigma \le \mbox{1/2.} \\
 \end{array}} \right.
\end{equation}
Due to the idea of Littlewood~\cite{2}, which is the only way of proving the
Lindel\"{o}f conjecture, the target of the paper is to give the proof of the
Lindel\"{o}f conjecture by the study of the
Lindel\"{o}f $\mu $ function (see~\cite{12}, p.18;~\cite{18}, p.338), with the aid of \textit{Theorem \ref{T1}},
which is proved based on the work of Heath-Brown~\cite{7}. The structure of the
paper is given as follows. In Section \ref{sec:2} we introduce the results for the
Riemann zeta function. In Section \ref{sec:3} we present the proof of
\textit{Conjecture \ref{C1}}. In Section \ref{sec:4} we prove the moment conjecture.
Finally, we suggest the new results on the moment for the Riemann zeta function
in Section \ref{sec:5}.

\section{Preliminaries} {\label{sec:2}}

In this section we give the recent results on the Riemann $\Xi$
and entire Riemann zeta functions.

Let $s_n $, $\varphi _n $ and $\phi _n $ run the nontrivial zeros of the
Riemann zeta function $\zeta \left( s \right)$, the imaginary part of the
Riemann zeta function $\zeta \left( s \right)$, and the positive imaginary
part of the Riemann zeta function $\zeta \left( s \right)$, respectively.

\begin{lemma}
\label{L1}

Let $s\in \mathcal{C}$ and $\wp =\log 2+\frac{1}{2}\log \pi -1-\frac{1}{2}\varpi
$, where $\varpi $ is the Euler's constant. Suppose that $\widetilde{s_n
}=Re\left( {s_n } \right)+\phi _n $, then the following representations
are equivalent:
\begin{equation}
\label{eq12}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
\begin{equation}
\label{eq13}
\xi \left( s \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1+\frac{i\left( {s-1/2} \right)}{\varphi _n }} \right)} ,
\end{equation}
\begin{equation}
\label{eq14}
\xi \left( s \right)=\xi \left( 0 \right)e^{s\wp }\prod\limits_{n=1}^\infty
{\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
\begin{equation}
\label{eq15}
\xi \left( s \right)=\xi \left( {1/2} \right)e^{s\wp
}\prod\limits_{l=1}^\infty {\left( {1+\frac{i\left( {s-1/2} \right)}{\varphi
_n }} \right)e^{s/\left( {1/2+i\varphi _n } \right)}} ,
\end{equation}
\begin{equation}
\label{eq16}
\xi \left( s \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1-\frac{\left( {s-1/2} \right)^2}{\phi _n^2 }} \right)} ,
\end{equation}
\begin{equation}
\label{eq17}
\xi \left( s \right)=\xi \left( {\frac{1}{2}}
\right)\prod\limits_{n=1}^\infty {\left[ {1-\frac{\left( {s-\frac{1}{2}}
\right)^2}{\left( {\widetilde{s_n }-\frac{1}{2}} \right)^2}} \right]} ,
\end{equation}
\begin{equation}
\label{eq18}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widetilde{s_n }}} \right)} \left( {1-\frac{s}{1-\widetilde{s_n
}}} \right),
\end{equation}
\begin{equation}
\label{eq19}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1-\frac{s\left( {1-s} \right)}{\widetilde{s_n }\left( {1-\widetilde{s_n }}
\right)}} \right]} ,
\end{equation}
and
\begin{equation}
\label{eq20}
\xi \left( s \right)=\sum\limits_{h=1}^\infty {\vartheta \left( h
\right)\left( {s-\frac{1}{2}} \right)^{2h}} ,
\end{equation}
where
\begin{equation}
\label{eq21}
\vartheta \left( h \right)=\frac{4}{\left( {2h}
\right)!}\int\limits_1^\infty {\frac{d\left( {z^{\frac{3}{2}}\psi ^{\left( 1
\right)}\left( x \right)} \right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log
x}{2}} \right)^{2h}dx} .
\end{equation}
\end{lemma}

\begin{proof}
See the details for the proof of \textit{Lemma \ref{L1}}~\cite{11}.
\end{proof}

\begin{remark}
The Hadamard product~(\ref{eq12}) was discovered by Hadamard in 1893~\cite{19}. Eq.~(\ref{eq13})
was discovered by Edwards~\cite{8} and proved by author in three ways~\cite{11}. Eq.~(\ref{eq14})
was discovered by Hadamard in 1893~\cite{19}, discussed by Landau in 1909
~\cite{20} and by Titchmarsh in 1930~\cite{21}, and proved in 1964 by Ingham~\cite{4}. Eq.~(\ref{eq15}),
discovered in~\cite{11}, was derived from Eq.~(\ref{eq14}). Eq.~(\ref{eq16}) was obtained
by Eq.~(\ref{eq17}) based on the Riemann theorem~\cite{11}. Eq.~(\ref{eq18}), derived from the
Patterson product (see~\cite{22}, p.34), e.g.,
\[
2\zeta \left( s \right)\Pi \left( s \right)=\prod\limits_{n=1}^\infty
{\left( {1-\frac{s}{\widetilde{s_n }}} \right)} \left(
{1-\frac{s}{1-\widetilde{s_n }}} \right),
\]
where $s\in \mathcal{C}$, leads to the equivalences of Eqs.~(\ref{eq18}) and~(\ref{eq19}) by
author~\cite{11}. Both~(\ref{eq18}) and~(\ref{eq19}) can be connected, as shown by
Edwards in 1974~\cite{8}.
Eq.~(\ref{eq20}) was discovered by Edwards in 1974~\cite{8}. From Eq.~(\ref{eq14}) we see that
$\xi \left( s \right)$ is the entire function of order $1$~\cite{7}.
\end{remark}

\begin{lemma}
\label{L2}
(Tur\'{a}n inequalities~\cite{11,13,23,24})

Let $h\ge 0$. Then the Tur\'{a}n inequalities
\begin{equation}
\label{eq22}
\left( {\vartheta \left( h \right)} \right)^2-\left( {\frac{2h-1}{2h+1}}
\right)\vartheta \left( {h-1} \right)\vartheta \left( {h+1} \right)>0
\end{equation}
hold for any $h\in \mathcal{N}\cup \left\{ 0 \right\}$.
\end{lemma}

\begin{proof}
See the proof of \textit{Lemma \ref{L2}}~\cite{11,13,23,24}.
\end{proof}

\begin{lemma}
\label{L3}
(Hardy theorem~\cite{10,25})

The Hardy theorem states that the entire Riemann zeta function $\xi \left( s
\right)$ has infinitely many zeros.
\end{lemma}

\begin{proof}
For the details of the proof of \textit{Lemma \ref{L3}}, see~\cite{10,25}.
\end{proof}

\begin{lemma}
\label{L4}

The entire Riemann zeta function $\xi \left( s \right)$ has infinitely many
zeros $s_n \in \mathcal{C}$.
\end{lemma}

\begin{proof}
For the proof of \textit{Lemma \ref{L4}}, see~\cite{11}.
\end{proof}

\begin{remark}
\textit{Lemma \ref{L4}} can be derived from \textit{Lemmas \ref{L2} and \ref{L3}}.
\end{remark}

Let $s=\frac{1}{2}+it$ such that~\cite{20}
\begin{equation}
\label{eq23}
\xi \left( {\frac{1}{2}+it} \right)=\Xi \left( t \right),
\end{equation}
where $t\in \mathcal{C}$.

\begin{lemma}
\label{L5}

Let $s\in \mathcal{C}$ and $\wp =\log 2+\frac{1}{2}\log \pi -1-\frac{1}{2}\varpi
$, where $\varpi $ is the Euler's constant. Suppose that $\widetilde{s_n
}=Re\left( {s_n } \right)+\phi _n $, then the following representations
are equivalent:
\begin{equation}
\label{eq24}
\Xi \left( t \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{s_n }} \right)} ,
\end{equation}
\begin{equation}
\label{eq25}
\Xi \left( t \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1-\frac{t}{\varphi _n }} \right)} ,
\end{equation}
\begin{equation}
\label{eq26}
\Xi \left( t \right)=\xi \left( 0 \right)e^{\wp \left( {\frac{1}{2}+it}
\right)}\prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}+it}{s_n }}
\right)e^{\left( {1/2+it} \right)/s_n }} ,
\end{equation}
\begin{equation}
\label{eq27}
\Xi \left( t \right)=\xi \left( {1/2} \right)e^{\wp \left( {\frac{1}{2}+it}
\right)}\prod\limits_{l=1}^\infty {\left( {1-\frac{t}{\varphi _n }}
\right)e^{\left( {1/2+it} \right)/\left( {1/2+i\varphi _n } \right)}} ,
\end{equation}
\begin{equation}
\label{eq28}
\Xi \left( t \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1-\frac{t^2}{\phi _n^2 }} \right)} ,
\end{equation}
\begin{equation}
\label{eq29}
\Xi \left( t \right)=\xi \left( {\frac{1}{2}}
\right)\prod\limits_{n=1}^\infty {\left[ {1+\frac{t^2}{\left(
{\widetilde{s_n }-\frac{1}{2}} \right)^2}} \right]} ,
\end{equation}
\begin{equation}
\label{eq30}
\Xi \left( t \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{\widetilde{s_n }}} \right)} \left(
{1-\frac{\frac{1}{2}+it}{1-\widetilde{s_n }}} \right),
\end{equation}
\begin{equation}
\label{eq31}
\Xi \left( t \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1-\frac{\frac{1}{4}+t^2}{\widetilde{s_n }\left( {1-\widetilde{s_n }}
\right)}} \right]} ,
\end{equation}
and
\begin{equation}
\label{eq32}
\Xi \left( t \right)=\sum\limits_{h=1}^\infty {\vartheta \left( h
\right)\left( {-1} \right)^ht^{2h}} ,
\end{equation}
where
\begin{equation}
\label{eq33}
\vartheta \left( h \right)=\frac{4}{\left( {2h}
\right)!}\int\limits_1^\infty {\frac{d\left( {z^{\frac{3}{2}}\psi ^{\left( 1
\right)}\left( x \right)} \right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log
x}{2}} \right)^{2h}dx} .
\end{equation}
\end{lemma}

\begin{proof}
For the details for the proof of \textit{Lemma \ref{L5}}, see~\cite{11}.
\end{proof}

\begin{lemma}
\label{L6}
The Riemann $\Xi$ function $\Xi \left( t \right)$ with $t\in \mathcal{
C}$ has infinitely many real zeros $\varphi _n \in \mathcal{R}$.
\end{lemma}

\begin{proof}
For the proof of \textit{Lemma \ref{L6}}, see~\cite{11}.
\end{proof}

\begin{remark}
It is well known that if $\vartheta \left( h \right)$ is given, then
\begin{equation}
\label{eq34}
\sum\limits_{h=1}^\infty {\vartheta \left( h \right)\left( {-1}
\right)^ht^{2h}} =\xi \left( {1/2} \right)e^{\hbar _0 \left(
{\frac{1}{2}+it} \right)}\prod\limits_{l=1}^\infty {\left(
{1-\frac{t}{\varphi _n }} \right)e^{\left( {1/2+it} \right)/\left(
{1/2+i\varphi _n } \right)}}
\end{equation}
is in the Laguerre-P\'{o}lya class~\cite{26,27}.

Eq.~(\ref{eq24}) was derived from the Hadamard product by author~\cite{11}. Eq.~(\ref{eq25}),
discovered in~\cite{11}, was derived from Eq.~(\ref{eq24}) based on the Riemann theorem.
As shown in~\cite{11}, Eq.~(\ref{eq26}) was derived from the Hadamard product~(\ref{eq14}) when
one takes $s=1/2+it$ in Eq.~(\ref{eq14}). Eq.~(\ref{eq27}), discovered in~\cite{11}, was derived
from Eq.~(\ref{eq26}) with the aid of Eqs.~(\ref{eq24}) and~(\ref{eq25}). Eq.~(\ref{eq28}), reported in 1894
by Cahen~\cite{28} and further discussed in 1927 by Titchmarsh~\cite{29}, was deduced
by the formula (here take the principal value)
\[
\log \Xi \left( t \right)=\log \xi \left( {1/2}
\right)+\sum\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\phi _n^2 }} \right)}
,
\]
which implies that [5]
\begin{equation}
\label{eq35}
\log \Xi \left( t \right)=\log \xi \left( {1/2}
\right)+\sum\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\phi _n^2 }} \right)}
,
\end{equation}
where
\begin{equation}
\label{eq36}
\xi \left( {1/2} \right)=\Xi \left( 0 \right)>0.
\end{equation}
Eq.~(\ref{eq28}) is in the case of Eq.~(\ref{eq29}) when the Riemann theorem is valid. Eq.~(\ref{eq29})
was derived by Eq.~(\ref{eq17}), which was deduced by the Patterson product
~\cite{22}. Eqs.~(\ref{eq30}) and~(\ref{eq31}), discovered in~\cite{11}, were equivalent to Eq.~(\ref{eq29})
since \textit{Lemma \ref{L1}} is valid for$s=1/2+it$. Eq.~(\ref{eq32}) was proposed by
Hadamard~\cite{19}, further studied by Jensen in 1913~\cite{30}, and discussed by
P\'{o}lya and Schur in 1914~\cite{27} and by P\'{o}lya in 1927~\cite{31}. It is easily seen
that $\Xi \left( t \right)$ is the even entire function of order $1$~\cite{10},
and that both $\varphi _n $ and $\phi _n $ are imaginary parts, obtained by
the Riemann-Siegel formula~\cite{32}. Here, the imaginary parts $\varphi _n $ and
$\phi _n $ are called as the Riemann-Siegel zeros.
\end{remark}

\begin{lemma}
\label{L7}
(Montgomery and Vaughan~\cite{18})

Let $\aleph >0$ be fixed. Then
\begin{equation}
\label{eq37}
\Im _1 t^{\frac{1}{2}-\sigma }\left| {\zeta \left( {1-s} \right)} \right|\le
\left| {\zeta \left( s \right)} \right|\le \Im _2 t^{\frac{1}{2}-\sigma
}\left| {\zeta \left( {1-s} \right)} \right|
\end{equation}
uniformly for $\left| \sigma \right|\le \aleph $, some positive absolute
constants $\Im _1 >0$ and $\Im _2 >0$, and $\left| t \right|\ge 1$.
\end{lemma}

\begin{proof}
See the result of Montgomery and Vaughan (see~\cite{18}, p.330).
\end{proof}

\begin{lemma}
\label{L8}

Suppose that $\mu \left( \sigma \right)$ is the Lindel\"{o}f $\mu $
function. Then,
\begin{equation}
\label{eq38}
\mu \left( \sigma \right)=\left\{ {\begin{array}{l}
 0,\mbox{ for }\sigma \ge \mbox{1/2,} \\
 1/2-\sigma ,\mbox{ for }\sigma \le \mbox{1/2.} \\
 \end{array}} \right.
\end{equation}
\end{lemma}

\begin{proof}
According to Hardy and Riesz (see~\cite{12}, p.18) and Montgomery and Vaughan (see~\cite{18}, p.338), we set
\begin{equation}
\label{eq39}
\Re \left( T \right)=\max _{0\le t\le T} \left| {\zeta \left( {1/2+it}
\right)} \right|.
\end{equation}
If $\Re \left( T \right)\ll T^\varepsilon $, then (see~\cite{18}, p.338)
\begin{equation}
\label{eq40}
\mu \left( \sigma \right)=0
\end{equation}
for $\sigma \ge 1/2$.

By \textit{Lemma \ref{L7}} it is shown that if $\Re \left( T \right)\ll
T^\varepsilon $, then (see~\cite{18}, p.338)
\begin{equation}
\label{eq41}
\mu \left( \sigma \right)=\frac{1}{2}-\sigma
\end{equation}
for $\sigma \le 1/2$.
\end{proof}

\section{The Lindel\"{o}f conjecture is true } {\label{sec:3}}

In this section we prove the Lindel\"{o}f conjecture by using the
Heath-Brown statement of the Riemann theorem.

\subsection{The new proof of the Riemann theorem}

We now consider the new proof of \textit{Theorem \ref{T1}}.

From the Hadamard product~\cite{19}
\begin{equation}
\label{eq42}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
and the Patterson product (see~\cite{22}, p.34)
\begin{equation}
\label{eq43}
2\xi \left( s \right)=\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widetilde{s_n }}} \right)} \left( {1-\frac{s}{1-\widetilde{s_n
}}} \right),
\end{equation}
we have~\cite{11}
\begin{equation}
\label{eq44}
\begin{array}{l}
 \xi \left( s \right)\\
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} \\
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widetilde{s_n }}} \right)} \left( {1-\frac{s}{1-\widetilde{s_n
}}} \right) \\
 =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left[
{1-\frac{\left( {s-\frac{1}{2}} \right)^2}{\left( {\widetilde{s_n
}-\frac{1}{2}} \right)^2}} \right]} , \\
 \end{array}
\end{equation}
where $\xi \left( 0 \right)=1/2$, and for $s=1/2+it$ we suggest~\cite{11}
\begin{equation}
\label{eq45}
\begin{array}{l}
 \Xi \left( t \right)\\
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{s_n }} \right)} \\
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{\widetilde{s_n }}} \right)} \left(
{1-\frac{\frac{1}{2}+it}{1-\widetilde{s_n }}} \right) \\
 =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{t^2}{\left( {\widetilde{s_n }-\frac{1}{2}} \right)^2}} \right]} \\
 =\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[ {1+\frac{t^2}{\left(
{\widetilde{s_n }-\frac{1}{2}} \right)^2}} \right]} , \\
 \end{array},
\end{equation}
where $t\in \mathcal{C}$ and
\[
\Xi \left( 0 \right)=\xi \left( {\frac{1}{2}} \right)>0.
\]
By using \textit{Lemma \ref{L5}} we have from Eq.~(\ref{eq45}) that
\begin{equation}
\label{eq46}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{t^2}{\left( {\widetilde{s_n }-\frac{1}{2}} \right)^2}} \right]}
=\sum\limits_{h=1}^\infty {\vartheta \left( h \right)\left( {-1}
\right)^ht^{2h}} ,
\end{equation}
where
\begin{equation}
\label{eq47}
\vartheta \left( h \right)=\frac{4}{\left( {2h}
\right)!}\int\limits_1^\infty {\frac{d\left( {z^{\frac{3}{2}}\psi ^{\left( 1
\right)}\left( x \right)} \right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log
x}{2}} \right)^{2h}dx} .
\end{equation}
By \textit{Lemma \ref{L6}} and the fact $\phi _n $ run the Riemann-Siegel zeros, we
have
\begin{equation}
\label{eq48}
\Xi \left( {\phi _n } \right)=0
\end{equation}
such that
\begin{equation}
\label{eq49}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{\phi _n^2 }{\left( {\widetilde{s_n }-\frac{1}{2}} \right)^2}}
\right]} =0
\end{equation}
since by \textit{Lemma \ref{L2}}, the Tur\'{a}n inequalities
\begin{equation}
\label{eq50}
\left( {\vartheta \left( h \right)} \right)^2-\left( {\frac{2h-1}{2h+1}}
\right)\vartheta \left( {h-1} \right)\vartheta \left( {h+1} \right)>0
\end{equation}
hold for any $h\in \mathcal{N}\cup \left\{ 0 \right\}$,
and by \textit{Lemma \ref{L3}}, the Hardy theorem is valid.

From Eq.~(\ref{eq49}) we find that
\begin{equation}
\label{eq51}
1+\frac{\phi _n^2 }{\left( {\widetilde{s_n }-\frac{1}{2}} \right)^2}\ne 0
\end{equation}
in which $\Xi \left( 0 \right)>0$.

Thus, from Eq.~(\ref{eq51}) we have
\begin{equation}
\label{eq52}
\widetilde{s_n }=\frac{1}{2}\pm i\phi _n
\end{equation}
and we rewrite Eq.~(\ref{eq44}) as
\begin{equation}
\label{eq53}
\xi \left( s \right)=\xi \left( {\frac{1}{2}}
\right)\prod\limits_{n=1}^\infty {\left[ {1+\frac{\left( {s-\frac{1}{2}}
\right)^2}{\phi _n^2 }} \right]}
\end{equation}
where $\xi \left( {\frac{1}{2}} \right)>0$ and $t\in \mathcal{C}$.

In view of \textit{Lemma \ref{L4}} we show that there exist $\widetilde{s_n }=Re\left(
{\widetilde{s_n }} \right)+i\phi _n $ such that
\begin{equation}
\label{eq54}
\xi \left( {\widetilde{s_n }} \right)=\xi \left( {Re\left( {\widetilde{s_n
}} \right)+i\phi _n } \right)=0.
\end{equation}

Combining Eqs.~(\ref{eq53}) and~(\ref{eq54}) we have
\begin{equation}
\label{eq55}
1+\frac{\left( {Re\left( {\widetilde{s_n }} \right)+i\phi _n -\frac{1}{2}}
\right)^2}{\phi _n^2 }=0,
\end{equation}
such that
\begin{equation}
\label{eq56}
\left[ {\left( {Re\left( {\widetilde{s_n }} \right)+i\phi _n -\frac{1}{2}}
\right)-i\phi _n } \right]\left[ {\left( {Re\left( {s_n } \right)+i\varphi
_n -\frac{1}{2}} \right)+i\phi _n } \right]=0.
\end{equation}

Because for $\phi _n \ne 0$ there exist
\begin{equation}
\label{eq57}
\left( {Re\left( {s_n } \right)+i\varphi _n -\frac{1}{2}} \right)+i\phi _n
\ne 0,
\end{equation}
we have
\begin{equation}
\label{eq58}
\left( {Re\left( {\widetilde{s_n }} \right)+i\phi _n -\frac{1}{2}}
\right)-i\phi _n =0.
\end{equation}

Thus,
\begin{equation}
\label{eq59}
Re\left( {\widetilde{s_n }} \right)=\frac{1}{2},
\end{equation}
which is the same as Eq.~(\ref{eq52}).

This implies that we prove the Heath-Brown statement~\cite{7} and that the zeros
for $\xi \left( s \right)$ are sample~\cite{7}.

\begin{figure}
  \centering
  \includegraphics[width=8cm]{fig1.eps}
  \caption{All zeros for the entire Riemann zeta function $\xi \left( s \right)$  lie on the critical line $Re\left( s \right)=1/2$ and in the critical strip $0<Re\left( s \right)<1$.}\label{fig1}
\end{figure}

Hence,
\begin{equation}
\label{eq60}
\zeta \left( {1/2+it} \right)=\zeta \left( {1/2+i\phi _n } \right)=0,
\end{equation}
in other worlds that the Riemann theorem follows.

All zeros for the entire Riemann zeta function $\xi \left( s \right)$ in the whole complex plane $s$ are demonstrated in Fig. \ref{fig1}.


\subsection{The proof of the Lindel\"{o}f conjecture}

In order to prove the Lindel\"{o}f conjecture, from Eq.~(\ref{eq38}) we have (see~\cite{18},
p.338)
\begin{equation}
\label{eq61}
\Re \left( T \right)=\max _{0\le t\le T} \left| {\zeta \left( {1/2+it}
\right)} \right|.
\end{equation}
Suppose that $\Re \left( T \right)\ll T^\varepsilon $ for any positive
$\varepsilon >0$ and $T\to \infty $ (see~\cite{18}, p.338).
By \textit{Lemma \ref{L8}}
we arrive at
\begin{equation}
\label{eq62}
\mu \left( \sigma \right)=0
\end{equation}
for $\sigma \ge 1/2$, and we give
\begin{equation}
\label{eq63}
\mu \left( \sigma \right)=\frac{1}{2}-\sigma
\end{equation}
for $\sigma \le 1/2$.

Since Eq.~(\ref{eq60}) is valid, we have
\begin{equation}
\label{eq64}
\mu \left( {\frac{1}{2}} \right)=\frac{1}{2}-\frac{1}{2}=0
\end{equation}
such that
\begin{equation}
\label{eq65}
\Re \left( T \right)\ll T^\varepsilon ,
\end{equation}
for $T\to \infty $ and any positive $\varepsilon >0$.

Combining Eqs.~(\ref{eq61}) and~(\ref{eq65}) we show that
\begin{equation}
\label{eq66}
\zeta \left( {1/2+iT} \right)\le \max _{0\le t\le T} \left| {\zeta \left(
{1/2+it} \right)} \right|\ll T^\varepsilon
\end{equation}
for $T\to \infty $ and any positive $\varepsilon >0$.

For $T\to \infty $ and any positive $\varepsilon >0$ we obtain
\begin{equation}
\label{eq67}
\zeta \left( {1/2+iT} \right)\ll T^\varepsilon .
\end{equation}
Thus, we finish the proof of the Lindel\"{o}f conjecture.

\begin{remark}
It has been pointed that the Riemann theorem implies that Lindel\"{o}f
conjecture~\cite{2,12,22,33}.
\end{remark}

\section{The moment conjecture is true}{\label{sec:4}}

In this section we give the proof of the moment conjecture based on the
Lindel\"{o}f conjecture.

According to Montgomery and Vaughan~\cite{18} the alternative representation for the Lindel\"{o}f conjecture
states that there is
\begin{equation}
\label{eq68}
\left| {\zeta \left( {1/2+it} \right)} \right|<\ell t^\varepsilon ,
\end{equation}
in which $\ell >0$ is an absolute constant, $t\to \infty $ and any positive
$\varepsilon >0$

By Eq.~(\ref{eq68}) and $m\in \mathcal{N}$ we have
\begin{equation}
\label{eq69}
\left| {\zeta \left( {1/2+it} \right)} \right|^m=\left| {\zeta ^m\left(
{1/2+it} \right)} \right|<\ell ^mt^{m\varepsilon },
\end{equation}
where $\ell >0$ is an absolute constant, $t\to \infty $ and any positive
$\varepsilon >0$

Making use of Eq.~(\ref{eq69}) there exist for every positive $\varepsilon _1 =m\varepsilon >0$ and $t\to \infty $,
\begin{equation}
\label{eq70}
\zeta ^m\left( {\frac{1}{2}\mbox{+}it} \right)\ll t^{\varepsilon _1 },
\end{equation}
which is in accordance with the result of Landau~\cite{34}.

With Eq.~(\ref{eq69}) we present
\begin{equation}
\label{eq71}
\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)} \right|^mdt}
=\int\limits_0^T {\left| {\zeta ^m\left( {1/2+it} \right)} \right|dt}
<\int\limits_0^T {\ell ^mt^{m\varepsilon }dt} =\frac{\ell ^m}{m\varepsilon
}T^{m\varepsilon +1},
\end{equation}
and
\begin{equation}
\label{eq72}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)}
\right|^mdt} =\frac{1}{T}\int\limits_0^T {\left| {\zeta ^m\left( {1/2+it}
\right)} \right|dt} <\frac{1}{T}\int\limits_0^T {\ell ^mt^{m\varepsilon }dt}
=\frac{\ell ^m}{m\varepsilon }T^{m\varepsilon +1},
\end{equation}
where $\ell >0$ is an absolute constant, $T\to \infty $ and any positive
$\varepsilon >0$.

From Eq.~(\ref{eq71}) we get
\begin{equation}
\label{eq73}
\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)} \right|^mdt} \ll
t^{\varepsilon _1 +1},
\end{equation}
and from Eq.~(\ref{eq72}) we have
\begin{equation}
\label{eq74}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)}
\right|^mdt} \ll T^{\varepsilon _1 },
\end{equation}
if $T\to \infty $ and $\varepsilon _1 =m\varepsilon >0$.

Similarly, with Eq.~(\ref{eq68})  we deduce
\begin{equation}
\label{eq75}
\left| {\zeta \left( {1/2+it} \right)} \right|^{2m}=\left| {\zeta
^{2m}\left( {1/2+it} \right)} \right|<\ell ^{2m}t^{2m\varepsilon },
\end{equation}
in which $\ell >0$ is an absolute constant, $m\in \mathcal{N}$, $t\to \infty $ and
any positive $\varepsilon >0$.

With use of Eq.~(\ref{eq75}) there exist
\begin{equation}
\label{eq76}
\zeta ^{2m}\left( {\frac{1}{2}\mbox{+}it} \right)\ll t^{\varepsilon _3 }
\end{equation}
for every positive $\varepsilon _3 =2\varepsilon _1 >0$, $m\in \mathcal{N}$ and
$t\to \infty $.

By Eq.~(\ref{eq75}) we give
\begin{equation}
\label{eq77}
\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)} \right|^{2m}dt}
=\int\limits_0^T {\left| {\zeta ^{2m}\left( {1/2+it} \right)} \right|dt}
<\int\limits_0^T {\ell ^{2m}t^{2m\varepsilon }dt} =\frac{\ell
^{2m}}{2m\varepsilon }T^{2m\varepsilon +1},
\end{equation}
and
\begin{equation}
\label{eq78}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)}
\right|^{2m}dt} =\frac{1}{T}\int\limits_0^T {\left| {\zeta ^{2m}\left(
{1/2+it} \right)} \right|dt} <\frac{1}{T}\int\limits_0^T {\ell
^{2m}t^{2m\varepsilon }dt} =\frac{\ell ^{2m}}{2m\varepsilon
}T^{2m\varepsilon }.
\end{equation}
From Eq.~(\ref{eq77}) we have for any positive $\varepsilon _4 =2m\varepsilon >0$
and $T\to \infty $,
\begin{equation}
\label{eq79}
\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)} \right|^{2m}dt} \ll
T^{\varepsilon _4 +1},
\end{equation}
which is in agreement with the result of Laurincikas and Steuding~\cite{14}.

By virtue of Eq.~(\ref{eq78}) we present
\begin{equation}
\label{eq80}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {1/2+it} \right)}
\right|^{2m}dt} \ll T^{\varepsilon _4 },
\end{equation}
which is in accord with the result~\cite{15}.

Thus, we prove the moment conjecture.

\section{New results on the moment for the Riemann zeta function}{\label{sec:5}}

In this section we propose the new results related to
the moment conjecture for the Riemann zeta function.
\begin{theorem}

\label{T2}

Let $s=\sigma +it\in \mathcal{C}$ with $\sigma ,t\in \mathcal{R}$, and $m\in \mathcal{N}$. Then
\begin{equation}
\label{eq81}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^mdt} \ll T^{\varepsilon _1 },
\end{equation}
or, alternatively,
\begin{equation}
\label{eq82}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^mdt} \ll
T^{\varepsilon _1 +1},
\end{equation}
where $T\to \infty $ and any positive $\varepsilon _1 >0$.
\end{theorem}

\begin{proof}
Since Eq. (\ref{eq5}) is valid, there exist an absolute constant $\beta >0$, $t\to
\infty $ and any positive $\varepsilon >0$ such that
\begin{equation}
\label{eq83}
\left| {\zeta \left( {\sigma +it} \right)} \right|<\beta t^\varepsilon ,
\end{equation}
where $\sigma ,t\in \mathcal{R}$.

For $m\in {\rm N}$ we have from Eq. (\ref{eq83}) that
\begin{equation}
\label{eq84}
\left| {\zeta \left( {\sigma +it} \right)} \right|^m=\left| {\zeta ^m\left(
{\sigma +it} \right)} \right|<\beta ^mt^{m\varepsilon }=\beta
^mt^{\varepsilon _1 },
\end{equation}
where $t\to \infty $.

This implies that
\begin{equation}
\label{eq85}
\zeta ^m\left( {\sigma +it} \right)\ll t^{\varepsilon _1 },
\end{equation}
in which $\varepsilon _1 =m\varepsilon >0$.

In view of Eq. (\ref{eq84}) we have
\begin{equation}
\label{eq86}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^mdt}
=\int\limits_0^T {\left| {\zeta ^m\left( {\sigma +it} \right)} \right|dt}
<\int\limits_0^T {\beta ^mt^{m\varepsilon }dt} =\frac{\beta
^mT^{m\varepsilon +1}}{m\varepsilon }=\frac{\beta ^m}{\varepsilon _1
}T^{\varepsilon _1 +1},
\end{equation}
which leads to
\begin{equation}
\label{eq87}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^mdt} \ll
T^{\varepsilon _1 +1},
\end{equation}
where $\varepsilon _1 =m\varepsilon >0$ and $T\to \infty $.

From Eq. (\ref{eq86}) we get
\begin{equation}
\label{eq88}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^mdt} =\frac{1}{T}\int\limits_0^T {\left| {\zeta ^m\left( {\sigma
+it} \right)} \right|dt} \ll \frac{1}{T}\int\limits_0^T {\beta
^mt^{m\varepsilon }dt} =\frac{1}{T}\frac{\beta ^mT^{m\varepsilon
+1}}{m\varepsilon }=\frac{\beta ^m}{\varepsilon _1 }T^{\varepsilon _1 },
\end{equation}
which yields that
\begin{equation}
\label{eq89}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^mdt} \ll T^{\varepsilon _1 },
\end{equation}
where $\varepsilon _1 =m\varepsilon >0$ and $T\to \infty $.
\end{proof}

\begin{theorem}

\label{T3}

Let $s=\sigma +it\in \mathcal{C}$ with $\sigma ,t\in \mathcal{R}$, and $m\in \mathcal{N}$. Then
\begin{equation}
\label{eq90}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^{2m}dt} \ll T^{\varepsilon _3 },
\end{equation}
or, alternatively,
\begin{equation}
\label{eq91}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^{2m}dt}
\ll T^{\varepsilon _3 +1},
\end{equation}
where $T\to \infty $.
\end{theorem}

\begin{proof}
For $m\in \mathcal{N}$ we deduce from Eq. (\ref{eq83}) that
\begin{equation}
\label{eq92}
\left| {\zeta \left( {\sigma +it} \right)} \right|^{2m}=\left| {\zeta
^{2m}\left( {\sigma +it} \right)} \right|<\beta ^{2m}t^{2m\varepsilon
}=\beta ^{2m}t^{\varepsilon _3 },
\end{equation}
where $t\to \infty $.

From Eq. (\ref{eq92}) we arrive at
\begin{equation}
\label{eq93}
\zeta ^{2m}\left( {\sigma +it} \right)\ll t^{\varepsilon _3 },
\end{equation}
where $\varepsilon _3 =2m\varepsilon >0$ and $t\to \infty $.

With use of Eq. (\ref{eq93}) we have
\begin{equation}
\label{eq94}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^{2m}dt}
=\int\limits_0^T {\left| {\zeta ^{2m}\left( {\sigma +it} \right)} \right|dt}
<\int\limits_0^T {\beta ^mt^{2m\varepsilon }dt} =\frac{\beta
^mT^{2m\varepsilon +1}}{2m\varepsilon }=\frac{\beta ^m}{\varepsilon _3
}T^{\varepsilon _3 +1},
\end{equation}
which implies that
\begin{equation}
\label{eq95}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^{2m}dt}
\ll T^{\varepsilon _3 +1},
\end{equation}
where $\varepsilon _3 =2m\varepsilon >0$ and $T\to \infty $.

By Eq. (\ref{eq92}) we show
\begin{equation}
\label{eq96}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^{2m}dt} =\frac{1}{T}\int\limits_0^T {\left| {\zeta ^{2m}\left(
{\sigma +it} \right)} \right|dt} \ll \frac{1}{T}\int\limits_0^T {\beta
^mt^{2m\varepsilon }dt} =\frac{1}{T}\frac{\beta ^mT^{2m\varepsilon
+1}}{2m\varepsilon }=\frac{\beta ^m}{\varepsilon _3 }T^{\varepsilon _3 },
\end{equation}
which yields that
\begin{equation}
\label{eq97}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^{2m}dt} \ll T^{\varepsilon _3 },
\end{equation}
where $\varepsilon _3 =2m\varepsilon >0$ and $T\to \infty $.
\end{proof}

\begin{remark}
For $m\in \mathcal{N}$ the following representations for the inequalities
are equivalent:
\begin{equation}
\label{eq18}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^mdt} \ll T^{\varepsilon _1 },
\end{equation}
\begin{equation}
\label{eq19}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^mdt} \ll
T^{\varepsilon _1 +1},
\end{equation}
where $\varepsilon _1 =m\varepsilon >0$ and $T\to \infty $, and
\begin{equation}
\label{eq20}
\zeta ^m\left( {\sigma +it} \right)\ll t^{\varepsilon _1 },
\end{equation}
where $t\to \infty $.

For $m\in \mathcal{N}$ the following representations for the inequalities
are equivalent:
\begin{equation}
\label{eq21}
\frac{1}{T}\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)}
\right|^{2m}dt} \ll T^{\varepsilon _3 },
\end{equation}
\begin{equation}
\label{eq22}
\int\limits_0^T {\left| {\zeta \left( {\sigma +it} \right)} \right|^{2m}dt}
\ll T^{\varepsilon _3 +1},
\end{equation}
where $\varepsilon _3 =2m\varepsilon >0$ and $T\to \infty $, and
\begin{equation}
\label{eq23}
\zeta ^m\left( {\sigma +it} \right)\ll t^{\varepsilon _3 },
\end{equation}
where $t\to \infty $.

It is easily seen that the equivalences for Conjecture \ref{C1} are true, as shown in~\cite{3,14,17}.
\end{remark}

\begin{acknowledgements}
This work is supported by the Yue-Qi Scholar of the China University of Mining and
Technology (No. 102504180004).
\end{acknowledgements}

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\end{document}
