\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{mathrsfs}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 232, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/232\hfil Remarks on regularity]
{Remarks on regularity criteria for the 3D Navier-Stokes equations}

\author[R. Wei, Y. Li  \hfil EJDE-2014/232\hfilneg]
{Ruiying Wei, Yin Li }  % in alphabetical order

\address{Ruiying Wei \newline
School of Mathematics and Information Science,
Shaoguan University,
Shaoguan, \newline Guangdong 512005, China.\newline
Department of Mathematics, Sun Yat-sen University ,
 Guangzhou, Guangdong 510275, China}
\email{weiruiying521@163.com}

\address{Yin Li \newline
School of Mathematics and Information Science,
Shaoguan University,
Shaoguan, \newline Guangdong 512005, China.\newline
Department of Mathematics, Sun Yat-sen University ,
 Guangzhou, Guangdong 510275, China}
\email{liyin2009521@163.com} 

\thanks{Submitted June 23, 2014. Published October 29, 2014.}
\subjclass[2000]{35Q35, 76D05}
\keywords{Regularity criteria; Triebel-Lizorkin space;
 Morrey-Campanato space}

\begin{abstract}
 In this article, we study the regularity criteria for the 3D Navier-Stokes
 equations involving derivatives of the partial components of the velocity.
 It is proved that if $\nabla_{h}\widetilde{u}$ belongs to Triebel-Lizorkin space,
 $\nabla u_3$ or $ u_3$ belongs to Morrey-Campanato space, then the
 solution remains smooth on $[0,T]$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

This article is devoted to the Cauchy problem for the following 
incompressible 3D Navier-Stokes  equation:
\begin{equation}  \label{e1.1}
\begin{gathered}
u_{t}+(u\cdot\nabla)u+\nabla p=\triangle u, \quad x\in \mathbb{R}^3,\;t>0 \\
 \operatorname{div} u=0, \quad x\in \mathbb{R}^3,\;t>0
\end{gathered}
\end{equation}
with initial data
\begin{equation} \label{e1.2}
u(x, 0)=u_{0},\quad x\in \mathbb{R}^3,
\end{equation}
where $u=(u_1(x,t),u_2(x,t),u_3(x,t))$ and $p=p(x,t)$  denote
the unknown velocity vector and
the unknown scalar pressure, respectively. In the last century, Leray
\cite{L} and Hopf \cite{H} proved the global existence of a weak solution
$u(x,t)\in L^{\infty}(0,\infty; 
L^2(\mathbb{R}^3))\cap L^2(0,\infty; H^1(\mathbb{R}^3))$
to \eqref{e1.1}-\eqref{e1.3} for any given initial datum 
$u_{0}(x)\in L^2(\mathbb{R}^3)$.
However, whether or not such a weak solution is regular and unique is still 
a challenging open problem.
From that time on, different criteria for regularity of the
weak solutions has been proposed.

The classical Prodi-Serrin conditions (see \cite{TO,GP,JS}) say  that if
$$
u\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad 
\frac{2}{t}+\frac{3}{s}=1,\quad 3\leq s\leq\infty,
$$
then the solution is smooth. Similar results is showed by Beir\~ao
 da Veiga \cite{HB} involving
the velocity gradient growth condition:
$$
\nabla u\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad 
\frac{2}{t}+\frac{3}{s}=2,\quad \frac{3}{2}<s\leq\infty.
$$

Actually, whether the weak solution is smooth when a part of the velocity 
components is involved. As for this direction,
later on, criteria just for one velocity component appeared. 
The first result in this direction is due to Neustupa et al \cite{N} 
(see also Zhou \cite{Z}), where the authors showed that if
$$  
u_3\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad
\frac{2}{t}+\frac{3}{s}=\frac{1}{2},\quad s\in(6,\infty],
$$
then the solution is smooth. A similar result, for the gradient of one velocity 
component, is independently due to Zhou \cite{ZY} and
Pokorn\'y \cite{P}. In \cite{ZY}, Zhou proved that if
$$ 
\nabla u_3\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad
\frac{2}{t}+\frac{3}{s}=\frac{3}{2},\quad 3\leq s<\infty.
$$
then the solution is smooth on $[0,T]$. This result is extended by Zhou and 
Pokorn\'y \cite{ZYM}; that is,
$$ 
\nabla u_3\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad
\frac{2}{t}+\frac{3}{s}=\frac{23}{12},\quad 2\leq s\leq 3.
$$
Further criteria, including several components of the velocity gradient, pressure
or other quantities, can be found, here we just list some. 
Zhou and Pokorn\'y \cite{ZP}  proved the 
regularity condition
$$ 
u_3\in L^{t}(0,T; L^{s}(\mathbb{R}^3)),\quad
\frac{2}{t}+\frac{3}{s}=\frac{3}{4}+\frac{1}{2s},\quad s>\frac{10}{3}.
$$
And in \cite{JZ}, Jia and Zhou proved that if  a weak solution $u$
 satisfies one of the following two conditions:
$$
 u_3\in L^{\infty}(0,T; L^{\frac{10}{3}}(\mathbb{R}^3));\quad
 \nabla u_3\in L^{\infty}(0,T; L^{30/19}(\mathbb{R}^3)),
$$
then $u$ is regular on $[0, T]$.
Dong and Zhang \cite{DZ} proved that if the horizontal derivatives
of the two velocity components
$$ 
\int_{0}^{T}\|\nabla_{h}\tilde{u}(s)\|_{\dot{B}^{0}_{\infty,\infty}}ds<\infty,
$$
then the solution keeps smoothness
up to time $T$, where $\tilde{u}=(u_1,u_2,0)$,
and $\nabla_{h}\tilde{u}=(\partial_1\tilde{u},\partial_2\tilde{u},0)$.
For other kinds of regularity criteria, see 
\cite{CT,FQ,FJ,JJ,ZG,ZG2,ZZ,ZA,ZYL} and the references cited therein.

Throughout this paper $C$ will denote a generic positive constant which can 
vary from line to line. For simplicity, we shall use
$\int f(x)\,dx$ to denote $\int_{R^3}f(x)\,dx$, use $\|\cdot\|_{p}$ to
denote $\|\cdot\|_{L^{p}}$.

The purpose of this article is to improve and extend above known regularity 
criterion of weak solution
for the equations \eqref{e1.1}, \eqref{e1.2} to the Triebel-Lizorkin space
 and Morrey-Campanato spaces.
The main results of this paper read:

\begin{theorem}\label{thm1.1}
Assume that $u_{0}\in H^1(\mathbb{R}^3)$ with $\operatorname{div}u_{0}=0$.
 u(x,t) is the corresponding weak solution to \eqref{e1.1} and \eqref{e1.2} 
on $[0,T)$. If additionally
\begin{equation} \label{e1.3}
\int_{0}^{T}\|\nabla_{h}\widetilde{u}(\cdot,t)\|^{p}_{\dot{F}^{0}_{q,\frac{2}{3}q}}dt
<\infty, \quad \text{with } \frac{2}{p}+\frac{3}{q}=2, \;\frac{3}{2}<q\leq\infty,
\end{equation}
then the solution remains smooth on $[0, T]$.
\end{theorem}

\begin{theorem}\label{thm1.2}
Assume that $u_{0}\in H^1(\mathbb{R}^3)$ with $\operatorname{div}u_{0}=0$.
$u(x,t)$ is the corresponding weak solution to \eqref{e1.1} and \eqref{e1.2} 
on $[0,T)$. If additionally
\begin{equation} \label{e1.4}
\int_{0}^{T}\|\nabla u_3(\cdot,t)\|^{\frac{8}{3(2-r)}}_{\dot{M}
^{p,\frac{3}{r}}}dt<\infty, \quad \text{with}\quad 0<r\leq 1,\; 2\leq p\leq\frac{3}{r},
\end{equation}
then the solution remains smooth on $[0, T]$.
\end{theorem}

\begin{theorem}\label{thm1.3}
Assume that $u_{0}\in H^1(\mathbb{R}^3)$ with $\operatorname{div}u_{0}=0$.
 u(x,t) is the corresponding weak solution to \eqref{e1.1} and \eqref{e1.2} 
on $[0,T)$. If additionally
\begin{equation} \label{e1.5}
\int_{0}^{T}\| u_3(\cdot,t)\|^{\frac{8}{3-4r}}_{\dot{M}^{p,\frac{3}{r}}}dt<\infty,
\quad \text{with } 0<r<\frac{3}{4},\; 2\leq p\leq\frac{3}{r},
\end{equation}
then the solution remains smooth on $[0, T]$.
\end{theorem}

\begin{remark} \label{rmk1.4}\rm
 Noticing that the classical Riesz transformation is bounded in 
$\dot{B}^{0}_{\infty,\infty}$,  if we take $q=\infty$ in Theorem \ref{thm1.1},  
then the classical Beal-Kato-Majda criterion for the Navier-Stokes equations 
is obtained; that is, if
$$
\int_{0}^{T}\|\nabla_{h}\widetilde{u}(\cdot,t)\|_{\dot{B}^{0}_{\infty,\infty}}dt
<\infty
$$
then the solution remains smooth on $[0, T]$.
\end{remark}

\begin{remark} \label{rmk1.5}\rm
 Since (it is proved in \cite{PG1,PG2})
\begin{gather*}
L^{q}(\mathbb{R}^3)=\dot M^{q,q}(\mathbb{R}^3)\subset
 \dot M^{p,q}(\mathbb{R}^3), \quad 1<p\leq q<\infty, \\
L^{\frac{3}{r}}(\mathbb{R}^3)\subset \dot{M}^{p,\frac{3}{r}}(\mathbb{R}^3)
\subset \dot X_{r}(\mathbb{R}^3)\subset \dot{M}^{2,\frac{3}{r}}(\mathbb{R}^3),
\quad  2<p\leq\frac{3}{r},\; 0<r<\frac{3}{2},
\end{gather*}
 the result of Theorem \ref{thm1.2} is an improved version of
\cite[Theorem 2]{ZZJ}. Also  we obtain, if
\[
\int_{0}^{T}\| u_3(\cdot,t)\|^{\frac{8}{3-4r}}_{\dot{X}_{r}}dt<\infty,
 \quad \text{with}\quad 0<r< \frac{3}{4},
\]
then the solution remains smooth on $[0, T]$.
\end{remark}


\section{Preliminaries}

In this section, we shall introduce the Littlewood-Paley decomposition theory, 
and then give some definitions of the homogeneous Besov space, 
homogeneous Triebel-Lizorkin space,
Morrey-Campanato space and multiplier space as well as some relate spaces used 
throughout this paper.
Before this, let us first recall the weak solutions of \eqref{e1.1}-\eqref{e1.3}:

 Let $u_{0}\in L^2(\mathbb{R}^3)$ with  $\nabla\cdot u_{0}=0$, a measurable
 $\mathbb{R}^3$-valued vector $u$ is said to be a weak solution of 
\eqref{e1.1}-\eqref{e1.3}  if the following conditions hold:
\begin{itemize}
\item[(1)] $u(x,t)\in L^{\infty}(0,\infty; L^2(\mathbb{R}^3))
 \cap L^2(0,\infty; H^1(\mathbb{R}^3))$;

\item[(2)] $u$ solves \eqref{e1.1}-\eqref{e1.2} in the sense of distributions;

\item[(3)]  the energy inequality holds; i.e,
$$
\| u\|^2_2+2\int_{0}^{t}\|\triangle u(\cdot,\tau)\|^2_2d\tau
\leq\| u_{0}\|^2_2,\ \ 0\leq t\leq T.
$$
\end{itemize}
Let us choose a nonnegative radial function $\varphi \in C^{\infty}(\mathbb{R}^3)$
be supported in the annulus 
$\{\xi\in \mathbb{R}^3: \frac{3}{4}\leq |\xi|\leq \frac{8}{3}\}$, such that
 $\sum^{\infty}_{l=-\infty}\varphi(2^{-l}\xi)=1, \forall \xi\neq0$.
For $f\in S'(\mathbb{R}^3)$, the frequency projection operators 
$\triangle_{l}$ is defined as
$$
\triangle_{l}f=\mathscr{F}^{-1}(\varphi(2^{-j}\cdot))\ast f,
$$
where $\mathscr{F}^{-1}(g)$ is the inverse Fourier transform of $g$.
 The formal decomposition
\begin{equation} \label{e2.1}
f=\sum^{\infty}_{l=-\infty}\triangle_{l}f.
\end{equation}
is called the homogeneous Littlewood-Paley decomposition. Noticing
$$
\triangle_{l}f=\sum^{l+1}_{j=l-1}\triangle_{j}(\triangle_{l}f)
$$
and  using the Young inequality, we have the following class Bernstein inequality:

\begin{lemma}[\cite{C}] \label{lem2.1}
Let $\alpha\in N $, then for all $1\leq p\leq q\leq\infty$,
$\sup_{|\alpha|=k}\|\partial^{\alpha}\triangle_{l}f\|_{q}
\leq C2^{lk+3l(\frac{1}{p}-\frac{1}{q})}\|\triangle_{l}f\|_{p}$.
 and $C$ is a constant independent of $f, l$.
\end{lemma}

For $s\in \mathbb{R}$ and $(p,q)\in [1,\infty]\times [1,\infty]$, the homogeneous 
Besov space $\dot{B}^{s}_{p,q}$  is defined by
$$
\dot{S}^{s}_{p,q}=\{f\in Z'(R^3): \|f\|_{\dot{S}^{s}_{p,q}}<\infty\},
$$
where
\[
\|f\|_{\dot{B}^{s}_{p,q}}=\begin{cases}
 \Big(\sum_{j \in z}2^{jsq}\|\triangle_{j}f(\cdot)\|_{p}^{q}\Big)^{1/q}, 
& 1\leq q<\infty,\\
\sup_{j\in z}2^{js}\|\triangle_{j}f(\cdot)\|_{p}, & q=\infty.
\end{cases}
\]
and $Z'(\mathbb{R}^3)$ denote the dual space of
$$
Z'(\mathbb{R}^3)=\{f\in \emph{S}(\mathbb{R}^3):D^{\alpha}\hat{f}(0)=0. \;
 \forall a\in N^3\}.
$$
On the other hand, for $s\in \mathbb{R}, (p,q)\in [1,\infty)\times [1,\infty]$,
 and for $s\in \mathbb{R}, p=q=\infty$, the homogenous Triebel-Lizorkin space is defined as
\begin{gather*}
\dot{F}^{s}_{p,q}=\{f\in Z'(\mathbb{R}^3): \|f\|_{\dot{F}^{s}_{p,q}}<\infty\},
\\
\|f\|_{\dot{F}^{s}_{p,q}} =\begin{cases}
\|(\sum_{j \in z}2^{jsq}|\triangle_{j}f(\cdot)|^{q})^{1/q}\|_{p}, & 1\leq q<\infty,\\
\|\sup_{j\in z}(2^{js}|\triangle_{j}f(\cdot)|)\|_{p}, & \ q=\infty.
\end{cases}
\end{gather*}
Notice that by Minkowski inequality, we have the following two imbedding relations:
\begin{gather*}
\dot{B}^{s}_{p,q}\subset\dot{F}^{s}_{p,q}, \quad q\leq p;\\
\dot{F}^{s}_{p,q}\subset\dot{B}^{s}_{p,q}, \quad p\leq q.
\end{gather*}
and the following two inclusions:
$$
\dot{H}^{s}=\dot{B}^{s}_{2,2}=\dot{F}^{s}_{2,2},\quad
L^{\infty}\subset\dot{F}^{0}_{\infty,\infty}=\dot{B}^{0}_{\infty,\infty}.
$$
We  refer to \cite{HT} for more properties.

For $1<q\leq p<\infty $, the homogeneous Morrey-Campanato space in 
$ \mathbb{R}^3$ is 
\[
\dot M^{p,q}=\big\{f\in L^{q}_{\rm loc}(\mathbb{R}^3);
\|f\|_{\dot M^{p,q}}=\sup _{x\in \mathbb{R}^3}\sup _{R>0}R^{\frac{3}{p}
-\frac{3}{q}}\|f\|_{{q}(B(x,R))}<\infty
\big\},
\]
For  $1\leq p'\leq q'<\infty $, we define the homogeneous space 
\begin{align*}
\dot N^{p',q'}=\Big\{&f\in L^{q'} | f=\sum_{k\in N}g_{k},
\text{ where $g_{k}\in L^{q'}_{comp}(R^3)$ and} \\
&\sum_{k\in N}d^{3(\frac{1}{p'}-\frac{1}{q'})}_{k}\|g_{k}\|_{{q'}}<\infty,
 \text{ where } d_{k}=\operatorname{diam}(\operatorname{supp}g_{k})<\infty
\Big\}.
\end{align*}
For $0<\alpha<3/2$, we say that a function belongs to the multiplier spaces 
$M(\dot H^{\alpha}, L^2)$ if it maps, by pointwise multiplication,
$\dot H^{\alpha}$ to $L^2$:
\[
\dot X_{\alpha}:=M(\dot H^{\alpha}, L^2)
:=\big\{f\in S'; \|f\cdot g\|_{L^2}\leq C\|g\|_{\dot H^{\alpha}} ,
\forall g\in \dot H^{\alpha}\big\}.
\]
Here, $\dot H^{\alpha}$ is the homogeneous Sobolev space of order $\alpha$,
\[
\dot H^{\alpha}=\big\{f\in L^1_{\rm loc}; \|f\|_{L^2}\equiv
\Big(\int _{R^3}|\xi|^{2\alpha}|\hat{u}(\xi)|^2\Big)^{1/2}<\infty\big\}.
\]
where $L^{p}(1\leq p\leq \infty)$ is the Lebsgue space endowed with norm 
$\|\cdot\|_{p}$.

\begin{lemma}[\cite{ SD, PG1}] \label{lem2.2}
Let $1\leq p'\leq q'<\infty $, and $p, q$ such that
$\frac{1}{p}+\frac{1}{p'}=1, \frac{1}{q}+\frac{1}{q'}=1$. 
Then $\dot M^{p,q}$ is the dual space of $\dot N^{p',q'}$.
\end{lemma}

\begin{lemma}[\cite{SD,FJ,PG1}] \label{lem2.3}
Let $1<p'\leq q'<2, m\geq2$, and $\frac{1}{p}+\frac{1}{p'}=1$. 
Denote $\alpha=-\frac{n}{2}+\frac{n}{p}+\frac{n}{m}\in(0, 1]$
Then there exists a constant $C>0$, such that for any 
$u\in L^{m}(\mathbb{R}^{n}), v\in \dot{H}^{\alpha}(\mathbb{R}^{n})$,
\[
\|u\cdot v\|_{\dot N^{p',q'}}\leq C\|u\|_{L^{m}}\| v\|_{\dot H^{\alpha}}.
\]
\end{lemma}

\begin{lemma}[\cite{PG2}] \label{lem2.4}
For $0\leq r\leq\frac{3}{2}$,  let the space 
$\mathscr{M}(\dot{B}_2^{r,1}\to L^2)$ be the space of functions which
 are locally square integrable
on $\mathbb{R}^3$ and such that pointwise multiplication with these functions
 maps boundedly the Besov space $\dot{B}_2^{r,1}(\mathbb{R}^3)$ to
$L^2(\mathbb{R}^3)$. The norm in $\mathscr{M}(\dot{B}_2^{r,1}\to L^2)$
is given by the operator norm of pointwise multiplication:
$$
\|f\|_{\mathscr{M}(\dot{B}_2^{r,1}\to L^2)}
=\sup\{\|fg\|_{{2}}:\|g\|_{\dot{B}_2^{r,1}}\leq1\}.
$$
Then, $f$ belongs to $\mathscr{M}(\dot{B}_2^{r,1}\to L^2)$ if and only
if $f$ belongs to $\dot M^{2,\frac{3}{r}}$ (with equivalence of norms).
\end{lemma}

\section{The proof of main results}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
Multiplying \eqref{e1.1}$_1$ by $-\triangle u$, integrating by parts,
noting that $\nabla\cdot u=0$, we have
\begin{equation} \label{e3.1}
\frac{1}{2}\frac{d}{dt}\int|\nabla u|^2\,dx+\int|\triangle u|^2\,dx
=\int[(u\cdot\bigtriangledown)u]\cdot\triangle u\,dx=:I.
\end{equation}
Next we estimate the right-hand side of \eqref{e3.1}, with the help of 
integration by parts and
$-\partial_3u_3=\partial_1u_1+\partial_2u_2$, one shows that
\begin{align*}
I&=-\sum^3_{i,j,k=1}\int\partial_{k}u_{i}\partial_{i}u_{j}\partial_{k}u_{j}\,dx\\
& =-\sum^2_{i,j=1}\sum^3_{k=1}\int\partial_{k}u_{i}\partial_{i}u_{j}
 \partial_{k}u_{j}\,dx
-\sum^2_{i,k=1}\int\partial_{k}u_{i}\partial_{i}u_3\partial_{k}u_3\,dx\\
&\quad -\sum^2_{j,k=1}\int\partial_{k}u_3\partial_3u_{j}\partial_{k}u_{j}\,dx
 -\sum^3_{k=1}\int\partial_{k}u_3\partial_3u_3\partial_{k}u_3\,dx\\
&\quad -\sum^2_{i=1}\int\partial_3u_{i}\partial_{i}u_3\partial_3u_3\,dx
 -\sum^2_{j=1}\int\partial_3u_3\partial_3u_{j}\partial_3u_{j}\,dx\\
&\leq C\int|\nabla_{h}\widetilde{u}||\nabla u|^2\,dx.
\end{align*}
Thus, the above inequality implies 
\begin{equation} \label{e3.2}
\frac{1}{2}\frac{d}{dt}\|\nabla u\|_2^2+\|\triangle u\|_2^2
\leq C\int|\nabla_{h}\widetilde{u}||\nabla u|^2\,dx.
\end{equation}
Using the Littlewood-Paley decomposition \eqref{e2.1}, 
$\nabla_{h}\widetilde{u}$ can be written as
\[
\nabla_{h}\widetilde{u}=\sum_{j<-N}\triangle_{j}(\nabla_{h}\widetilde{u})
+\sum_{j=-N}^{N}\triangle_{j}(\nabla_{h}\widetilde{u})
+\sum_{j>N}\triangle_{j}(\nabla_{h}\widetilde{u}).
\]
where $N$ is a positive integer to be chosen later. 
Substituting this into \eqref{e3.2}, one obtains
\begin{equation} \label{e3.3}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\|\nabla u\|_2^2+\|\triangle u\|_2^2 \\
&\leq C\sum_{j<-N}\int|\triangle_{j}(\nabla_{h}\widetilde{u})||\nabla u|^2\,dx
 +C\sum_{j=-N}^{N}\int|\triangle_{j}(\nabla_{h}\widetilde{u})||\nabla u|^2\,dx\\
&\quad +C\sum_{j>N}\int|\triangle_{j}(\nabla_{h}\widetilde{u})||\nabla u|^2\,dx\\
&=:K_1+K_2+K_3.
\end{aligned}
\end{equation}
For $K_{i}$ $(i=1,2,3)$, we now give the estimates one by one. 
For $K_1$, using the H\"{o}lder inequality, the Young inequality and
 Lemma \ref{lem2.1}, it follows that
\begin{equation} \label{e3.4}
\begin{aligned}
K_1&\leq C\sum_{j<-N}\|\triangle_{j}(\nabla_{h}\widetilde{u})
 \|_{\infty}\|\nabla u\|^2_2\\
&\leq C\sum_{j<-N}2^{3j/2}\|\triangle_{j}(\nabla_{h}\widetilde{u})
 \|_2\|\nabla u\|^2_2\\
&\leq C\Big(\sum_{j<-N}2^{3j}\Big)^{1/2}
\Big(\sum_{j<-N}\|\triangle_{j}(\nabla_{h}\widetilde{u})\|_2^2\Big)^{1/2}
\|\nabla u\|^2_2\\
&\leq C2^{-3N/2}\|\nabla u\|^3_2.
\end{aligned}
\end{equation}
Where in the last inequality, we use the fact that for all 
$s\in \mathbb{R}$, $\dot{H}^{s}=\dot{B}^{s}_{2,2}$.

For $K_2$, by the H\"{o}lder inequality and the Young inequality,  one has
\begin{equation} \label{e3.5}
\begin{aligned}
K_2&=C\int\sum_{j=-N}^{N}|\triangle_{j}(\nabla_{h}\widetilde{u})||\nabla u|^2\,dx\\
&\leq CN^{\frac{2q-3}{2q}}\int\Big(\sum_{j=-N}^{N}|\triangle_{j}
(\nabla_{h}\widetilde{u})|^{2q/3}\Big)^{3/(2q)}|\nabla u|^2\,dx\\
&\quad \leq CN^{\frac{2q-3}{2q}}\|\nabla_{h}\widetilde{u}
 \|_{\dot{F}^{0}_{q,\frac{2q}{3}}}\|\nabla u\|^2_{\frac{2q}{q-1}}\\
&\quad \leq CN^{\frac{2q-3}{2q}}\|\nabla_{h}\widetilde{u}
 \|_{\dot{F}^{0}_{q,\frac{2q}{3}}}\|\nabla u\|^{\frac{2q-3}{q}}_2
 \|\triangle u\|_2^{3/q}\\
&\quad \leq\frac{1}{2}\|\triangle u\|_2^2+CN\|\nabla_{h}\widetilde{u}
 \|_{\dot{F}^{0}_{q,\frac{2q}{3}}}^{\frac{2q}{2q-3}}\|\nabla u\|^2_2.
\end{aligned}
\end{equation}
where we used the  interpolation inequality
\[
\|u\|_{s}\leq C\|u\|_2^{\frac{3}{s}
-\frac{1}{2}}\|u\|_{\dot{H}^1}^{\frac{3}{2}-\frac{3}{s}},
\]
for $2\leq s\leq 6$.

Finally,  using  H\"{o}lder inequality and  Lemma \ref{lem2.1}, $K_3$ can be estimated as
\begin{equation} \label{e3.6}
\begin{aligned}
K_3
&=C\sum_{j>N}\int|\triangle_{j}(\nabla_{h}\widetilde{u})||\nabla u|^2\,dx\\
&\leq C\sum_{j>N}\|\triangle_{j}(\nabla_{h}\widetilde{u})\|_3\|\nabla u\|_3^2\\
&\leq C\sum_{j>N}2^{\frac{j}{2}}\|\triangle_{j}(\nabla_{h}\widetilde{u})
 \|_2\|\nabla u\|_3^2\\
&\leq C(\sum_{j>N}2^{-j})^{1/2}(\sum_{j>N}2^{2j}\|\triangle_{j}(\nabla_{h}
 \widetilde{u})\|_2^2)^{1/2}
\|\nabla u\|_2\|\triangle u\|_2\\
&\leq C2^{-N/2} \|\nabla u\|_2\|\triangle u\|_2^2.
\end{aligned}
\end{equation}
Substituting \eqref{e3.4}, \eqref{e3.5} and \eqref{e3.6} in \eqref{e3.3}, 
we obtain
\begin{equation} \label{e3.7}
\begin{aligned}
&\frac{d}{dt}\|\nabla u\|_2^2+\|\triangle u\|_2^2\\
&\leq C2^{-\frac{3}{2}N}\|\nabla u\|^3_2+CN\|\nabla_{h}\widetilde{u}
 \|_{\dot{F}^{0}_{q,\frac{2q}{3}}}^{\frac{2q}{2q-3}}\|\nabla u\|^2_2
 +C2^{-N/2}\|\nabla u\|_2\|\triangle u\|_2^2.
\end{aligned}
\end{equation}
Now we choose $N$ such that
$C2^{-N/2}\|\nabla u\|_2\leq \frac{1}{2}$; that is
$$
N\geq\frac{\ln(\|\nabla u\|_2^2+e)+\ln C}{\ln 2}+2.
$$
Thus \eqref{e3.7} implies
\begin{align*}
\frac{d}{dt}\|\nabla u\|_2^2
\leq C+C\|\nabla_{h}\widetilde{u}\|_{\dot{F}^{0}_{q,\frac{2q}{3}}}^{p}
\ln(\|\nabla u\|_2^2+e)\|\nabla u\|^2_2.
\end{align*}
Taking the Gronwall inequality into consideration, we obtain
\begin{align*}
\ln(\|\nabla u\|_2^2+e)
\leq C\Big[1+\int^{T}_{0}\|\nabla_{h}\widetilde{u}
\|_{\dot{F}^{0}_{q,\frac{2q}{3}}}^{p}(\tau)d\tau 
\cdot e^{\int^{T}_{0}\|\nabla_{h}\widetilde{u}\|_{\dot{F}^{0}_{q,\frac{2q}{3}}}^{p}
(\tau)d\tau}\Big].
\end{align*}
The proof of Theorem \ref{thm1.1} is complete under the condition \eqref{e1.3}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
Multiplying \eqref{e1.1}$_1$ by $-\triangle_{h} u$, integrating by parts,
 noting that $\nabla\cdot u=0$, we have
\begin{equation} \label{e3.8}
\frac{1}{2}\frac{d}{dt}\int|\nabla_{h} u|^2\,dx+\int|\nabla\nabla_{h} u|^2\,dx
=\int[(u\cdot\bigtriangledown)u]\cdot\triangle_{h} u\,dx=:J.\
\end{equation}
Next we estimate the right-hand side of \eqref{e3.8}, with the help of 
integration by parts and
$-\partial_3u_3=\partial_1u_1+\partial_2u_2$, one shows that
\begin{equation} \label{e3.9}
\begin{aligned}
J&=-\sum^3_{i,j=1}\sum^2_{k=1}\int\partial_{k}u_{i}\partial_{i}u_{j}
 \partial_{k}u_{j}\,dx\\
&=-\sum^2_{i,j,k=1}\int\partial_{k}u_{i}\partial_{i}u_{j}\partial_{k}u_{j}\,dx
-\sum^2_{i,k=1}\int\partial_{k}u_{i}\partial_{i}u_3\partial_{k}u_3\,dx\\
&\quad -\sum^3_{j=1}\sum^2_{k=1}\int\partial_{k}u_3\partial_3u_{j}\partial_{k}u_{j}\,dx\\
&=:J_1+J_2+J_3.
\end{aligned}
\end{equation}
For $J_2$  and $J_3$, we obtain
\begin{equation} \label{e3.10}
|J_2+J_3|\leq C\int|\nabla u_3||\nabla_{h}{u}||\nabla u|\,dx.
\end{equation}
$J_1$ is a sum of eight terms, using the fact 
$-\partial_3u_3=\partial_1u_1+\partial_2u_2$, we can estimate it as
\begin{equation} \label{e3.11}
\begin{aligned}
J_1&=-\int(\partial_1u_1+\partial_2u_2)[(\partial_1u_1)^2
-\partial_1u_1\partial_2u_2+(\partial_2u_2)^2]\,dx\\
&\quad -\int(\partial_1u_1+\partial_2u_2)[(\partial_2u_1)^2
 +\partial_1u_2\partial_2u_1+(\partial_1u_2)^2]\,dx\\
&=\int\partial_3u_3[(\partial_1u_1)^2-\partial_1u_1\partial_2u_2+(\partial_2u_2)^2+
(\partial_2u_1)^2+\partial_1u_2\partial_2u_1+(\partial_1u_2)^2]\,dx\\
&\leq C\int|\nabla u_3||\nabla_{h}{u}||\nabla u|\,dx.
\end{aligned}
\end{equation}
Substituting the estimates \eqref{e3.9}-\eqref{e3.11} in \eqref{e3.8}, we obtain
\begin{equation} \label{e3.12}
\frac{1}{2}\frac{d}{dt}\|\nabla_{h} u\|_2^2+\|\nabla\nabla_{h} u\|_2^2
\leq C\int|\nabla u_3||\nabla_{h}{u}||\nabla u|\,dx=:L.
\end{equation}
when $2<p\leq \frac{3}{r}$, using Lemmas \ref{lem2.2} and \ref{lem2.3},  
and the Young inequality, 
we obtain
\begin{equation} \label{e3.13}
\begin{aligned}
L&\leq C\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}\||\nabla u|\cdot|\nabla_{h} u
 |\|_{\dot N^{p',\frac{3}{3-r}}}\\
&\leq C\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}\|\nabla_{h}u\|_{\dot H^{r}}
 \|\nabla u\|_2\\
&\leq C\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}\|\nabla u\|_{L^2}
 \|\nabla_{h} u\|_2^{1-r}\|\nabla\nabla_{h} u\|_2^{r}\\
&\leq \frac{1}{2}\|\nabla\nabla_{h} u\|_2^2+C\|\nabla u_3
 \|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{2-r}}\|\nabla u\|_2^2.
\end{aligned}
\end{equation}
where we used the  inequality
\[
\|f\|_{\dot{H}^{r}}=\||\xi|^{r}\hat{f}\|_2
=(\int |\xi|^{2r}|\hat{f}|^{2r}|\hat{f}|^{2-2r}d\xi)^{1/2}
\leq \|f\|_2^{1-r}\|\nabla f\|_2^{r}, 
\]
with $ 0< r\leq 1$.

In the case  $p=2$, using H\"{o}lderĄ¯s inequality, Lemma \ref{lem2.4}, and 
the Young inequality, we can estimate $L$ as
\begin{equation} \label{e3.14}
\begin{aligned}
L&\leq C\||\nabla u_3|\cdot|\nabla_{h} u|\|_2\|\nabla u\|_2\\
&\leq C\|\nabla u_3\|_{\dot M^{2,\frac{3}{r}}}\|\nabla_{h}u
 \|_{\dot{B}^{r,1}_2}\|\nabla u\|_2\\
&\leq C\|\nabla u_3\|_{\dot M^{2,\frac{3}{r}}}\|\nabla_{h} u
 \|_2^{1-r}\|\nabla\nabla_{h} u\|_2^{r}\|\nabla u\|_2\\
&\leq \frac{1}{2}\|\nabla\nabla_{h} u\|_2^2+C\|\nabla u_3
 \|_{\dot M^{2,\frac{3}{r}}}^{\frac{2}{2-r}}\|\nabla u\|_2^2.
\end{aligned}
\end{equation}
where we used the following interpolation inequality \cite{SM}: 
for $0\leq r\leq1$, 
$\|f\|_{\dot{B}^{r,1}_2}\leq \|f\|_2^{1-r}\|\nabla f\|_2^{r}$.

Now, gathering \eqref{e3.13} and \eqref{e3.14} together and substituting 
into \eqref{e3.12}, we obtain
\begin{equation} \label{e3.15}
\frac{d}{dt}\|\nabla_{h} u\|_2^2+\|\nabla\nabla_{h} u\|_2^2
\leq C\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{2-r}}\|\nabla u\|_2^2.
\end{equation}
Multiplying \eqref{e1.1}$_1$ by $-\triangle u$, integrating by parts, 
noting that $\nabla\cdot u=0$, we have (see \cite{ZP})
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|\nabla u\|_2^2+\|\triangle u\|_2^2
&=\int[(u\cdot\bigtriangledown)u]\cdot\triangle u\,dx\\
&\leq C\int|\nabla_{h}u||\nabla u|^2\,dx\\
&\leq C\|\nabla_{h}u\|_2\|\nabla u\|^2_{4}\\
&\leq C\|\nabla_{h}u\|_2\|\nabla u\|^{1/2}_2\|\nabla u\|^{\frac{3}{2}}_{6}\\
&\leq C\|\nabla_{h}u\|_2\|\nabla u\|^{1/2}_2\|\nabla\nabla_{h} u
\|_2\|\triangle u\|^{1/2}_2.
\end{align*}
Integrating, with respect to $t$,  yields
\begin{equation} \label{e3.16}
\begin{aligned}
&\frac{1}{2}\|\nabla u(t)\|_2^2+\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\\
&\leq \frac{1}{2}\|\nabla u_{0}\|_2^2
 +C\sup _{0\leq \tau\leq t}\|\nabla_{h}u(\tau)\|_2(\int^{t}_{0}\|\nabla u(\tau)
 \|^2_2d\tau)^{1/4}\\
&\quad\times \Big(\int^{t}_{0}\|\nabla\nabla_{h} u(\tau)\|^2_2d\tau\Big)^{1/2}
\Big(\int^{t}_{0}\|\triangle u(\tau)\|^2_2d\tau\Big)^{1/4}.
\end{aligned}
\end{equation}
Substituting \eqref{e3.15} in \eqref{e3.16}, using H\"{o}lderĄ¯s inequality  
and the Young inequality, we obtain
\begin{equation} \label{e3.17}
\begin{aligned}
&\frac{1}{2}\|\nabla u(t)\|_2^2+\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\\
&\leq \frac{1}{2}\|\nabla u_{0}\|_2^2+(C+C\int^{t}_{0}\|\nabla 
 u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{2-r}}\|\nabla u(\tau)\|_2^2d\tau)
\Big(\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\Big)^{1/4}\\
&\leq C+C\Big(\int^{t}_{0}\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{2-r}}
 \|\nabla u(\tau)\|_2^{\frac{3}{2}}\|\nabla u(\tau)\|_2^{1/2}d\tau\Big)^{4/3}\\
&\quad +\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\\
&\leq C+C\Big(\int^{t}_{0}\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{8}{3(2-r)}}
 \|\nabla u(\tau)\|_2^2d\tau\Big)
\Big(\int^{t}_{0}\|\nabla u(\tau)\|_2^2d\tau\Big)^{1/3} \\
&\quad +\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\\
&\leq C+C\int^{t}_{0}\|\nabla u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{8}{3(2-r)}}
 \|\nabla u(\tau)\|_2^2d\tau+\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau.
\end{aligned}
\end{equation}
Absorbing the last term into the left hand side, applying the Gronwall inequality 
and combining with the standard continuation
argument, we conclude that the solutions $u$ can be extended
beyond $t = T$ provided that 
$\nabla u_3\in L^{\frac{8}{3(2-r)}}(0,T;\dot{M}^{p,\frac{3}{r}}(\mathbb{R}^3))$. 
This completes the proof of Theorem \ref{thm1.2}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.3}]
We start from \eqref{e3.9}, we can estimate $J_2$ and $J_3$ as
\begin{equation} \label{e3.18}
|J_2+J_3|\leq C\int|u_3||\nabla u||\nabla\nabla_{h}{u}|\,dx.
\end{equation}
From \eqref{e3.11}, we  find that
\begin{equation} \label{e3.19}
J_1\leq C\int|u_3||\nabla u||\nabla\nabla_{h}{u}|\,dx.
\end{equation}
Combining \eqref{e3.9},\eqref{e3.17}, \eqref{e3.18} with \eqref{e3.8}, we obtain
\begin{equation} \label{e3.20}
\frac{1}{2}\frac{d}{dt}\|\nabla_{h} u\|_2^2+\|\nabla\nabla_{h} u\|_2^2
\leq C\int|u_3||\nabla u||\nabla\nabla_{h}{u}|\,dx=:V.
\end{equation}
When $2<p\leq \frac{3}{r}$,  similarly as in the proof of $L$ in \eqref{e3.13},
 we obtain
\begin{equation} \label{e3.21}
\begin{aligned}
V&\leq C\| u_3\|_{\dot M^{p,\frac{3}{r}}}\||\nabla u|\cdot|\nabla\nabla_{h} 
 u|\|_{\dot N^{p',\frac{3}{3-r}}}\\
&\leq C\| u_3\|_{\dot M^{p,\frac{3}{r}}}\|\nabla u\|_{\dot H^{r}}\|\nabla\nabla_{h} 
 u\|_2\\
&\leq C\| u_3\|_{\dot M^{p,\frac{3}{r}}}\|\nabla \nabla_{h}u\|_{L^2}
 \|\nabla u\|_2^{1-r}\|\triangle u\|_2^{r}\\
&\leq \frac{1}{2}\|\nabla\nabla_{h} u\|_2^2+C\| u_3\|_{\dot M^{p,\frac{3}{r}}}^2
 \|\nabla u\|_2^{2(1-r)}\|\triangle u\|_2^{2r}.
\end{aligned}
\end{equation}
When $p=2$,  similarly as in the proof of $L$ in \eqref{e3.14}, we have
\begin{equation} \label{e3.22}
\begin{aligned}
V&\leq C\|| u_3|\cdot|\nabla u|\|_2\|\nabla\nabla_{h} u\|_2\\
&\leq C\| u_3\|_{\dot M^{2,\frac{3}{r}}}\|\nabla u\|_{\dot{B}^{r,1}_2}
 \|\nabla \nabla_{h}u\|_2\\
&\leq C\| u_3\|_{\dot M^{2,\frac{3}{r}}}\|\nabla u\|_2^{1-r}
 \|\triangle u\|_2^{r}\|\nabla\nabla_{h} u\|_2\\
&\leq \frac{1}{2}\|\nabla\nabla_{h} u\|_2^2+C\| u_3
 \|_{\dot M^{2,\frac{3}{r}}}^2\|\nabla u\|_2^{2(1-r)}
\|\triangle u\|_2^{2r}.
\end{aligned}
\end{equation}
Substituting \eqref{e3.21} and  \eqref{e3.22} in  \eqref{e3.20}, we  find that
\begin{equation} \label{e3.23}
\frac{d}{dt}\|\nabla_{h} u\|_2^2+\|\nabla\nabla_{h} u\|_2^2
\leq C\| u_3\|_{\dot M^{2,\frac{3}{r}}}^2\|\nabla u\|_2^{2(1-r)}
\|\triangle u\|_2^{2r}.
\end{equation}
Substituting \eqref{e3.23} in \eqref{e3.16}, we obtain
\begin{align*}
&\frac{1}{2}\|\nabla u(t)\|_2^2+\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\\
&\leq \frac{1}{2}\|\nabla u_{0}\|_2^2+\Big(C+C\int^{t}_{0}
\| u_3\|_{\dot M^{p,\frac{3}{r}}}^2\|\nabla u(\tau)\|_2^{2(1-r)}
\|\triangle u(\tau)\|_2^{2r}d\tau\Big)\\
&\quad\times \Big(\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau\Big)^{1/4}\\
&\leq C+\frac{1}{4}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau
+C\Big(\int^{t}_{0}\| u_3\|_{\dot M^{p,\frac{3}{r}}}^2\|\nabla u(\tau)\|_2^{2(1-r)}
\|\triangle u(\tau)\|_2^{2r}d\tau\Big)^{4/3}\\
&\leq C+\frac{1}{4}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau
+C\Big(\int^{t}_{0}
\|\triangle u(\tau)\|_2^2d\tau\Big)^{4r/3}\\
&\quad\times \Big(\int^{t}_{0}\| u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{1-r}}
\|\nabla u(\tau)\|_2^2d\tau\Big)^{\frac{4(1-r)}{3}}\\
&\leq C+\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau
+C\Big(\int^{t}_{0}\| u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{2}{1-r}}
 \|\nabla u(\tau)\|_2^{\frac{3-4r}{2(1-r)}}\|\nabla u(\tau)
 \|_2^{\frac{1}{2(1-r)}}d\tau\Big)^{\frac{4(1-r)}{3-4r}}\\
&\leq C+\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau
 +C\Big(\int^{t}_{0}\| u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{8}{3-4r}}
\|\nabla u(\tau)\|_2^2d\tau\Big)
\Big(\int^{t}_{0}\|\nabla u(\tau)\|_2^2d\tau\Big)^{\frac{1}{3-4r}}\\
&\leq C+\frac{1}{2}\int^{t}_{0}\|\triangle u(\tau)\|_2^2d\tau
+C\int^{t}_{0}\| u_3\|_{\dot M^{p,\frac{3}{r}}}^{\frac{8}{3-4r}}
\|\nabla u(\tau)\|_2^2d\tau.
\end{align*}

By a similar argument as in the proof of Theorem \ref{thm1.2}, provided that
 $ u_3\in L^{\frac{8}{3-4r}}(0,T;\dot{M}^{p,\frac{3}{r}}(\mathbb{R}^3))$, 
we complete the proof of Theorem \ref{thm1.3}.
\end{proof}

\subsection*{Acknowledgements}
Ruiying Wei and Yin Li would like to express sincere gratitude to
Professor Zheng-an Yao of Sun Yat-sen University for enthusiastic guidance
 and constant encouragement.
This work is partially supported by Guangdong Provincial culture of seedling 
of China (no. 2013LYM0081), and Guangdong
Provincial NSF of China (no. S2012010010069), the Shaoguan
Science and Technology Foundation (no. 313140546), and
Science Foundation of Shaoguan University.

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\end{document}