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\markboth{\hfil smoothness of solutions to
  the Navier-Stokes equations \hfil EJDE--2003/11}
{EJDE--2003/11\hfil Milan Pokorn\'y \hfil}

\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2003}(2003), No. 11, pp. 1--8. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  On the result of He concerning the smoothness of solutions to
  the Navier-Stokes equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 35Q35, 76D05.
\hfil\break\indent
{\em Key words:} Navier-Stokes equations, regularity of
 systems of PDE's, \hfil\break\indent
 anisotropic regularity criteria.
\hfil\break\indent
\copyright 2003 Southwest Texas State University. \hfil\break\indent
Submitted November 6, 2002. Published February 6, 2003.
\hfil\break\indent
Partially supported by the Grant Agency of the Czech Republic
(grants 201/00/0768 \hfil\break\indent and 201/02/P091) 
and by the Council of the Czech Government (project 113200007)
} }
\date{}
%
\author{Milan Pokorn\'y}
\maketitle

\begin{abstract}
 We improve the regularity criterion for the Navier-Stokes equations
 proved  by He \cite{He}. We show that for the Cauchy problem the
 Leray-Hopf weak solution is smooth provided
 $\nabla u_3 \in L^t(0,T;L^s)$, $\frac 2t + \frac 3s \leq \frac 32$.
\end{abstract}

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

\section{Introduction and Main Theorem}

We consider the Cauchy problem for the Navier-Stokes equations in three space dimensions, i.e. the
system of PDE's
\begin{equation} \label{1}
\begin{array}{cl}
\left.\begin{array} {c} \displaystyle \varrho \frac{\partial \mathbf{u}}{\partial t}
+ \varrho (\mathbf{u} \cdot \nabla) \mathbf{u} - \nu \Delta \mathbf{u} + \nabla p = \varrho \mathbf{f}
\\[6pt]
\nabla \cdot \mathbf{u} = 0
\end{array} \right\} &\mbox{ in } (0,T)\times {\mathbb{R}}^3
\\[6pt]
\mathbf{u}(0,\mathbf{x}) = \mathbf{u}_0(\mathbf{x}) &\mbox{ in } {\mathbb{R}}^3\, .
\end{array}
\end{equation}

Here, $\mathbf{u}: (0,T)\times \mathbb{R}^3 \mapsto \mathbb{R}^3$ is the velocity field,
$p: (0,T) \times \mathbb{R}^3 \mapsto \mathbb{R}^3$ is the pressure,
$\mathbf{f}: (0,T)\times \mathbb{R}^3 \mapsto \mathbb{R}^3$
denotes
the volume force, $0<T<\infty$. For our purpose, the values of the constant density $\varrho$ and the constant viscosity $\nu$ do not play any role; we therefore assume without loss of generality $\varrho = \nu =1$. Moreover, in order to simplify the presentation
of the result, we take $\mathbf{f} = \mathbf{0}$.

As is well known, the existence of globally in time smooth solution to
system (\ref{1}) is proved only for small data \cite{La}; for
large
data we only have  the existence of a weak solution \cite{Le2},
which is locally in time smooth provided the data are smooth enough
\cite{KiLa}.

On the other hand, if we assume that our weak solution
is "slightly" smoother than it follows from
the definition then such a solution is as smooth as the data of the
 problem allow (provided the data are smooth enough).
We call $\mathbf{u} \in L^\infty(0,T; L^2) \cap L^2 (0,T; W^{1,2})$ with
$\nabla
\cdot \mathbf{u} = \mathbf{0}$ a weak solution to (\ref {1}) with $\mathbf{f} =\mathbf{0}$, if $\langle
\mathbf{u}',
\mathbf{v}\rangle + \int\nabla \mathbf{u} : \nabla \mathbf{v} + \int ((\mathbf{u}\cdot \nabla)\mathbf{u})
\cdot \mathbf{v} = 0$ for a.a. $t \in (0,T)$ and all $\mathbf{v} \in W^{1,2}$ with
 $\nabla \cdot \mathbf{v} = 0$, and $\lim_{t \to 0+} \mathbf{u}(t) = \mathbf{u}_0$ in the
weak $L^2$ sense.

 Let us mention some of these regularity criteria
\begin{itemize}
\item[(I)] $\mathbf{u} \in L^t(I;L^s)$, $\frac 2t + \frac 3s \leq 1$, $2\leq t \leq
\infty$, $3\leq s\leq \infty$ (see \cite {Se}, for the case $s=3$ see
\cite{SeSv}, \cite{ErSeSv})
\item[(II)] $\nabla\mathbf{u} \in L^t(I;L^s)$, $\frac 2t + \frac 3s \leq 2$, $1\leq t \leq
\infty$, $\frac 32 < s\leq \infty$ (see \cite {BV})
\item[(III)] $p \in   L^t(I;L^s)$, $\frac 2 t + \frac 3 s \leq 2$, $1\leq t \leq \infty$, $\frac 32 < s\leq \infty$ (see \cite{BeGa})
\end{itemize}

On the other hand, in two space dimensions the weak solution is known
to be unique and as regular as the data of the problem allow (see
\cite{Le1}). Therefore several authors tried to find regularity
criteria which depend only on one velocity component and/or on the
derivatives of one velocity component or derivatives only in the $x_3$
direction
\begin{itemize}
\item[(IV)] $u_3 \in L^t(I;L^s)$, $\frac 2t + \frac 3s \leq \frac 12$,
$4\leq t \leq
\infty$, $6<s\leq \infty$ (see
\cite{NeNoPe})
\item[(V)] $u_3 \in L^t(I;L^s)$, $\frac 2 t + \frac 3 s \leq 1$,
$2\leq t \leq \infty$, $3 < s\leq \infty$ and $\frac{\partial u_1}{\partial x_3}
, \frac{\partial u_2}{\partial x_3} \in L^t(I;L^s)$, 
$\frac 2 t + \frac 3 s \leq 2$,
$1\leq t \leq \infty$, $\frac 32 < s\leq \infty$
\item[(VI)] $\frac{\partial u_3}{\partial x_3} \in L^\infty(I;L^\infty)$
\item[(VII)] $\frac{\partial \mathbf{u}}{\partial x_3}\in L^t(I;L^s)$, $\frac 2 t 
+ \frac 3 s \leq \frac 32$,
$\frac 43\leq t \leq \infty$, $2 \leq s\leq \infty$
\item[(VIII)]  $\frac{\partial u_3}{\partial x_3}
 \in L^t(I;L^s)$, $\frac 2t + \frac 3s \leq 1$,
$2 \leq t \leq \infty$, $3 \leq s\leq \infty$ and
$\frac{\partial u_1}{\partial x_3}, \frac{\partial u_2}{\partial x_3}
\in L^t(I;L^s)$, $\frac 2 t + \frac 3 s \leq
2$,
$1\leq t \leq \infty$, $\frac 32 < s\leq \infty$
(For the results (V)--(VIII) see \cite{PePo}.)
\end{itemize}

In the recent paper \cite{He}, He followed similar aim and obtained the
regularity of the Navier-Stokes system provided $\nabla \mathbf{u}_3 \in
L^t(I;L^s)$, $\frac 2t + \frac 3s \leq 1$, $2\leq t\leq \infty$,
$3\leq s \leq \infty$. This result, in comparison to the result of
Neustupa, Novotn\'y and Penel \cite{NeNoPe}, does not seem to be
optimal. One would rather expect in this case $\frac 2t + \frac 3s \leq \frac
32$. The aim of this note is to show that this is indeed true. More
precisely

\begin{theorem} \label{t1}
Let $\mathbf{u}_0 \in W^{1,2}(\mathbb{R}^3)$ with ${\rm div}\, \mathbf{u}_0 = 0$,
$\mathbf{f} = \mathbf{0}$ and let $\mathbf{u}$ be a weak solution to the Navier-Stokes
equations (\ref{1}) which satisfies the energy inequality\footnote{We
say that a weak solution to the Navier-Stokes equations (\ref{1}) satisfies the
energy inequality if for almost all $t\in (0,T)$ it holds
$$
\frac 12 \frac{d}{dt} \Big(\int|\mathbf{u}|^2(t)\Big) + \int |\nabla \mathbf{u}|^2 (t) \leq 0\, .
$$
It is not difficult to show that such weak solutions exist; on the
other hand, it is not known whether any weak solution  in the
sense above satisfies the
energy inequality. Weak solutions satisfying the energy inequality are
usually called Leray-Hopf weak solutions}. Assume moreover that $\nabla u_3 \in L^t(0,T;L^s)$ with $\frac 2t + \frac 3s \leq \frac 32$, $\frac 43 \leq
t \leq \infty$, $2 \leq s \leq \infty$. Then $\mathbf{u}$ and the
corresponding pressure $p$ is the smooth
solution to the Navier-Stokes equations, i.e. $\mathbf{u} \in
L^\infty(0,T;W^{1,2}(\mathbb{R}^3) \cap L^{2}(0,T; W^{2,2}(\mathbb{R}^3))$,
$\nabla p \in L^{2}(0,T;
W^{1,2}(\mathbb{R}^3))$. Moreover $\mathbf{u} \in
C^\infty([\delta, T)\times \mathbb{R}^3)$ and $p \in
C^\infty([\delta, T)\times \mathbb{R}^3)$ for $delta$ any small positive
number.
\end{theorem}

\begin{remark} \label {r1} {\rm
Assuming $\mathbf{f} \neq \mathbf{0}$ we would get
the regularity of the solution in dependence on the regularity
$\mathbf{f}$. Since these calculations are relatively standard, we omit
them here.}
\end{remark}

\begin{remark} \label{r2} {\rm
At the first sight Theorem \ref{t1} seems
to be a direct consequence of the result from
\cite{NeNoPe}. But this is true only for $2\leq s<3$, i.e. for the
case when $W^{1,s}\hookrightarrow L^{\frac {3s}{3-s}}$. Nevertheless,
we will prove Theorem 1 also in this case.}
\end{remark}

\section{Proof of Theorem \ref{t1}}

In what follows, we use standard notation for the Sobolev and
Lebesgue spaces ($W^{k,p}$ and $L^p$, respectively) as well as for the
corresponding norms \hbox{($\|\,\cdot\,\|_{k,p}$} and $\|\,\cdot\,\|_{p}$, respectively) without specifying
the domain (always $\mathbb{R}^3$). Morerover, the Bochner spaces $L^p(I;X)$
will be in the case of $X=L^q$ denoted shortly $L^{p,q}$.
In order to simplify the notation, we will not distinguish between
$(L^p)^m$ and $L^p$.

Any generic constant will denoted by $C$; its value may vary, even on
the same line or in the same formula. We also use the summation
convention.

The proof will be a modification of the procedure used by Neustupa,
Novotn\'y and Penel (see \cite{NeNoPe} and also \cite{NePe}), where
regularity criteria only for suitable weak solutions were studied.
This proof can be also regarded as a way how
to transform the results from the above mentioned papers to the Cauchy
problem.

First, as $\mathbf{u}_0 \in W^{1,2}$ with $\mbox{div}\, \mathbf{u}_0 = 0$, we know
that there exists exactly one strong solution to the Navier-Stokes
equations with the initial condition $\mathbf{u}_0$ (on a possibly short time
interval). Denote
$$\tau_0 = \sup_{\tau>0} \Big\{\mbox{there exists a
strong solution to (\ref{1}) on } (0,\tau)\Big\}\, .
$$
It is well known that $\tau_0 >0$.
As our weak solution from Theorem \ref{t1} satisfies the energy inequality,
it coincides with the strong solution on its interval of existence
(see e.g. \cite{Te}). We will show that the assumption $\tau_0
<T$ leads to a contradiction. Note that the solution is smooth on
the open interval $(0,\tau_0)$ and thus the equations are satisfied
pointwise here.

Denote by $Y = L^\infty(0,\tau; L^2) \cap
L^2(0,\tau;W^{1,2})$ with $0<\tau <\tau_0$. We will not specify the
length of the time interval $(0,\tau)$ in the notation for $Y$. Our aim
will be to show that under the assumptions of Theorem \ref{t1},
$\nabla \mathbf{u}$ remains bounded in $Y$ independently of $\tau$, provided
$\tau_0<T$. Thus, using standard extension argument, we get a
contradiction with the maximality of $\tau_0$.

To this aim, we first
show that for any $\tau<\tau_0$
\begin{equation} \label {2}
\|\omega_3\|_Y^2 \equiv \|\omega_3\|_{L^\infty(0,\tau;L^2)}^2 + \|\nabla
\omega_3 \|_{L^2(0,\tau;L^2)}^2 \leq C_1 + C_2 \|\omega\|_Y
\end{equation}
with $C_i= C_i(\mathbf{u}_0,\|\nabla u_3\|_{L^{t,s}})$, $i=1,2$. In
particular, the constants are independent of $\tau$. (Here, by
$\omega$ we denote the vorticity, i.e. $\omega =\mbox{curl}\, \mathbf{u}$.)
Using (\ref{2})
it will be relatively easy to show that
$$
\|\nabla \mathbf{u}\|_Y \leq C (\mathbf{u}_0, \|\nabla u_3\|_{L^{s,t}})
$$
with the constant independent of $\tau$. This finishes our proof as
our weak solution cannot blow up at $\tau_0$.

Let us first prove (\ref{2}).

\begin{lemma} \label{l1}
Under the assumptions of Theorem \ref{t1}, there exist positive constants
$C_1(\mathbf{u}_0, \|\nabla \mathbf{u}\|_{L^{t,s}})$ and $C_2(\mathbf{u}_0, \|\nabla \mathbf{u}\|
_{L^{t,s}})$ such that (\ref{2}) holds true.
\end{lemma}

\noindent{\bf Proof:}
As explained above, it is enough to show inequality (\ref{2}) for
smooth solutions to (\ref{1}). To this aim, let us look at the
equation for the vorticity. We have
\begin{equation} \label{2a}
\frac{\partial \omega}{\partial t} -\Delta \omega + (\mathbf{u}\cdot\nabla)\omega - (\omega\cdot
\nabla) \mathbf{u} = \mathbf{0}\, .
\end{equation}

Multiplying the equation for $\omega_3$ by $\omega_3$ and integrating
over $\mathbb{R}^3$ we get (note that all integrals are finite)
\begin{equation} \label {3}
\frac 12 \frac{d}{dt} \|\omega_3\|_2^2 + \|\nabla \omega_3\|_2^2 =
\int (\omega \cdot \nabla)u_3  \omega_3 \equiv I_1\, .
\end{equation}
We will now estimate $I_1$. Using H\"older's inequality and standard
interpolation inequalities we have ($\frac 1p + \frac 1q + \frac 1s =
1$, $2\leq s \leq \infty$, $2\leq p\leq 6$, $2\leq q\leq 3$)
$$
\begin{array}{rl}
|I_1| &\leq \|\nabla u_3\|_s \|\omega_3\|_p \|\omega\|_q\\[5pt]
&\leq \|\nabla u_3\|_s \|\omega_3\|_2^{\frac{6-p}{2p}}
\|\omega_3\|_6^{\frac{3p-6}{2p}}
\|\omega\|_2^{\frac{6-q}{2q}}\|\omega\|_6^{\frac{3q-6}{2q}} \\[6pt]
&\leq \displaystyle \frac 12 \|\nabla \omega_3\|_2^2 + C \|\nabla u_3\|_s^{\frac{4p}{6+p}}
\|\omega\|_2^{2 \frac pq \frac{6-q}{6+p}} \|\omega\|_6^{2\frac pq
\frac{3q-6}{6+p}} \|\omega_3\|_2^{2 \frac {6-p}{6+p}}\, .
\end{array}
$$
Thus
$$
\frac {d}{dt} \|\omega_3\|_2^{\frac {4p}{6+p}} \leq C \|\nabla u_3\|
_s^{\frac{4p}{6+p}}
\|\omega\|_2^{2 \frac pq \frac{6-q}{6+p}} \|\omega\|_6^{2\frac pq
\frac{3q-6}{6+p}}\, ,
$$
i.e.
$$
\|\omega_3\|_2^{\frac{4p}{6+p}}(\tau) \leq \|\omega_3\|_2^{\frac
{4p}{6+p}}(0) + C \|\omega\|_{L^{\infty,2}}^{4\frac {p}{q}
(\frac{3-q}{6+p})}  \int_0^{\tau} \|\nabla u_3\|_s^{\frac{4p}{6+p}}
\|\omega\|_2^{\frac{2p}{6+p}} \|\omega\|_6^{2\frac pq
\frac{3q-6}{6+p}} ds\, .
$$
Now, $\frac{3s-4}{6s} + \frac{p}{6+p} + \frac{3q-6}{q}\frac{p}{6+p} =
1$ (recall that $\frac 1p + \frac 1q +\frac 1s =1$) and we get
$$
\|\omega_3\|_{L^\infty(0,\tau;L^2)}^{\frac{4p}{6+p}} \leq C(\mathbf{u}_0) + C \|\nabla u_3\|_{L^{t,s}}^{\frac{4p}{6+p}}
\|\omega\|_{L^{2,2}}^{\frac{2p}{6+p}} \|\omega\|_Y^{
\frac{2p}{6+p}}\, ,
$$
i.e.
\begin{equation} \label{4}
\|\omega_3\|_{L^{\infty,2}}^2 \leq C_1 + C_2 \|\omega\|_Y\, .
\end{equation}
Returning to (\ref{3}), repeating calculations above  and using (\ref{4}) we get the
desired inequality (\ref{2}). \hfill$\Box$ \smallskip

\noindent{\bf Proof of Theorem \ref{t1}:}
We rewrite equation (\ref{1})$_1$ in the form
\begin{equation} \label {5}
\frac{\partial \mathbf{u}}{\partial t} - \Delta \mathbf{u} + (\omega \times \mathbf{u}) + \nabla (p+ \frac 12
|\mathbf{u}|^2) = \mathbf{0}\, .
\end{equation}
Multiplying equation (\ref{5}) by $-\Delta \mathbf{u}$ and integrating over
$\mathbb{R}^3$ we easily see that for $0<\tau <\tau_0$
\begin{equation} \label {5a}
\frac 12 \frac{d}{dt} \|\nabla \mathbf{u}\|_2^2(\tau) + \int \|\nabla^2
\mathbf{u}\|_2^2 (\tau) = \int(\omega\times \mathbf{u}) \cdot \Delta \mathbf{u} \equiv I_2
\end{equation}
Using
$$
(\omega\times \mathbf{u})\cdot \Delta \mathbf{u} = (\omega_2 u_3-\omega_3 u_2)\Delta
u_1  + (\omega_3 u_1-\omega_1 u_3)\Delta u_2 + (\omega_1 u_2-\omega_2
u_1) \Delta u_3
$$
we get
$$
\begin{array}{rl}
\displaystyle
|I_2| &\leq C\Big(\int |\nabla \mathbf{u}|^2 |\nabla u_3| + |\mathbf{u}| |\nabla^2
 \mathbf{u}| |\nabla u_3| + \int |\mathbf{u}||\nabla^2 \mathbf{u}| |\omega_3|\Big) \\[6pt]
\displaystyle &\equiv
 C(I_{21} + I_{22} +I_{23})\, .
\end{array}
$$
We estimate each term separately; $I_{21}$ and $I_{22}$ using
better regularity properties of $\nabla u_3$, $I_{23}$ using Lemma
\ref{l1}.
If $s<\infty$,
\begin{equation} \label {6}
I_{21} \leq \|\nabla u_3\|_s \|\nabla \mathbf{u}\|_{\frac {2s}{s-1}}^2 \leq
\varepsilon \|\nabla ^2\mathbf{u}\|_2^2 + C(\varepsilon) \|\nabla
\mathbf{u}_3\|_s^{\frac{2s}{2s-3}} \|\nabla \mathbf{u}\|_2^2\, .
\end{equation}
Similarly we proceed for $s=\infty$.
\begin{equation} \label{8a}
I_{22} \leq \varepsilon \|\nabla^2 \mathbf{u}\|_2^2 + C(\varepsilon) \int
|\mathbf{u}|^2 |\nabla u_3|^2\, .
\end{equation}
If $2\leq s\leq 3$ we estimate the integral on the right-hand side by
$\|\nabla u_3\|_s^2 \|\mathbf{u}\|_{\frac {2s}{s-2}}^2$ and using the
interpolation inequality
$$
\|\mathbf{u}\|_{\frac{2s}{s-2}} \leq C \|\nabla \mathbf{u}\|_2^{\frac{2s-3}{s}}
\|\nabla^2 \mathbf{u}\|_{2}^{\frac{3-s}{s}}
$$
we get
\begin{equation} \label {7}
I_{22} \leq 2 \varepsilon \|\nabla^2 \mathbf{u}\|_2^2 + C(\varepsilon)
\|\nabla u_3\|_s^{\frac{2s}{2s-3}} \|\nabla\mathbf{u}\|_2^2\, .
\end{equation}
For $s>3$ we estimate
$$
\int |\mathbf{u}|^2 |\nabla u_3|^2 \leq \|\nabla u_3\|_s^{2(1-\alpha)}
\|\nabla u_3\|_2^{2\alpha} \|\mathbf{u}\|_{\frac{2s}{(s-2)(1-\alpha)}}^2\, ,
$$
where for our purpose the optimal choice of $\alpha$ is
$\frac{2s-6}{5s-6}$ (for $s<\infty$) and $s=\frac 25$ (for
$s=\infty$), respectively. Note that $0\leq \alpha \leq \frac
25$. Now, as $\frac {10}{3} \leq \frac{2s}{(s-2)(1-\alpha)}\leq 6$, we
use the interpolation inequality
$$
\|\mathbf{u}\|_{\frac 23 \frac{5s-6}{s-2}}^2 \leq C
\|\mathbf{u}\|_2^{4\frac{s-3}{5s-6}} \|\nabla \mathbf{u}\|_2^{\frac {6s}{5s-6}}
$$
and thus
$$
\int |\mathbf{u}|^2 |\nabla u_3|^2 \leq C \|\nabla
u_3\|_s^{\frac{6s}{5s-6}}\|\nabla \mathbf{u}\|_2^2 \|\mathbf{u}\|_2^{4\frac
{s-3}{5s-6}}\, .
$$
Taking (\ref{8a}) into account we end up with
$$
I_{22} \leq \varepsilon \|\nabla^2 \mathbf{u}\|_2^2 + C(\varepsilon, \mathbf{u}_0)
\|\nabla \mathbf{u}_3\|_s^{\frac {6s}{5s-6}} \|\nabla\mathbf{u}\|_2^2\,.
%\tag{\ref{7}'}
$$
Note that, even though $\frac{6s}{5s-6} \geq \frac{2s}{2s-3}$ for
$s\geq 3$, we still have with $t= \frac{6s}{5s-6}$ that $\frac 2t +
\frac 3s = \frac 53 + \frac 1s$ for $3\leq s \leq \infty$.

Finally we consider $I_{23}$. Here we apply Lemma \ref{l1} and
\begin{equation} \label{9}
I_{23} \leq \|\nabla^2 \mathbf{u}\|_2 \|\omega_3\|_3 \|\mathbf{u}\|_6 \leq
\varepsilon \|\nabla^2 \mathbf{u}\|_2^2 + \varepsilon \|\omega_3\|_3^4 +
C(\varepsilon) \|\nabla \mathbf{u}\|_2^4\, .
\end{equation}
Inequalities (\ref{5a})--(\ref{9}), after integrating over
$(0,\tau)$, $\tau < \tau_0$ read as follows
\begin{equation} \label{10}
\begin{array}{rl}
&\displaystyle \frac 12 \|\nabla \mathbf{u}\|_2^2 (\tau) + \int_0^\tau
 \|\nabla^2 \mathbf{u}\|_2^2 \\[6pt]
&\displaystyle \leq K\varepsilon  \int_0^\tau \|\nabla^2 \mathbf{u}\|_2^2 + \varepsilon
\int_0^\tau \|\omega_3\|_3^4 \\[6pt]
&\quad +\displaystyle C(\varepsilon,\mathbf{u}_0)\int_0^\tau(\|\nabla u_3\|_s^
{\frac{2s}{2s-3}} + g(s)\|\nabla u_3\|_s^{\frac {6s}{5s-6}} +
\|\nabla\mathbf{u}\|_2^2)\|\nabla \mathbf{u}\|_2^2\, ,
\end{array}
\end{equation}
where $g(s) = 0$  for $2\leq s\leq 3$ and $g(s) =1$ for $s>3$.
Lemma \ref{l1} yields
$$
\int_0^\tau \|\omega_3\|_3^4 \leq C_1(\mathbf{u}_0,\|\nabla u_3\|_{L^{t,s}}) +
C_2(\mathbf{u}_0, \|\nabla u_3\|_{L^{t,s}}) \|\nabla \mathbf{u}\|_Y^2\, .
$$
(Note that the larger $\|\nabla u_3\|_{L^{t,s}}$ is, the smaller
$\varepsilon$ must be.)
Thus from (\ref{10}), taking $\varepsilon$ sufficiently small, it
follows that
$$
\begin{array}{l}
{\displaystyle \|\nabla \mathbf{u}\|_2^2 (\tau) + \int_0^\tau
 \|\nabla^2\mathbf{u}\|_2^2  }\\[6pt]
\displaystyle \leq  \sup_{\sigma \in (0,\tau)} \|\nabla
 \mathbf{u}\|_2^2  (\sigma) + \int_0^\tau \|\nabla^2 \mathbf{u}\|_2^2 \\[6pt]
 \displaystyle \leq  C_1(\mathbf{u}_0,\|\nabla u_3\|_{L^{t,s}}) \\[6pt]
\displaystyle \quad +
C_2(\mathbf{u}_0, \|\nabla u_3\|_{L^{t,s}}) \int_0^\tau \big(\|\nabla
 u_3\|_s^{\frac{2s}{2s-3}} + g(s) \|\nabla u_3\|_s^{\frac{6s}{5s-6}} +
 \|\nabla \mathbf{u}\|_2^2\big) \|\nabla \mathbf{u}\|_2^2
\end{array}
$$
and, applying the Gronwall inequality, we obtain
$$
\|\nabla \mathbf{u}\|_{L^\infty(0,\tau;L^2)}^2 + \|\nabla^2
\mathbf{u}\|_{L^2(0,\tau;L^2)}^2 \leq C(\mathbf{u}_0,\|\nabla u_3\|_{L^{t,s}})\, ,
$$
where the constant $C$ is in particular independent of $\tau$ as $\tau
\to \tau_0$. Theorem \ref{t1} is proved.
\hfill$\Box$

\begin{remark} {\rm I would like to thank the referee who kindly informed me
that a similar result as presented in Theorem \ref{t1}
was recently obtained by Y. Zhou \cite{Zh}.
The main idea of the proof (i.e. the estimate of $\omega_3$ in
Lemma \ref{l1} and consequently of $\omega$ (proof of Theorem \ref{t1}))
is basically the same. On the other hand, the too papers differ in the
way how the quantities on the right-hand side are estimated as well as
in the argument how the formally obtained a priori estimates are verified
for only weak solutions to the Navier-Stokes equations.
I was also kindly informed by the authors below that a similar problem,
for $s=3$ and suitable weak solutions in bounded domains, was also
considered by Z. Skal\'ak and P. Ku\v{c}era in \cite{SkKu}. }
\end{remark}

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\noindent\textsc{Milan Pokorn\'{y}}\\
Math. Institute of Charles University\\
Sokolovsk\'a 83, 186 75 Praha 8, Czech Republic\\
e-mail: {\tt pokorny@karlin.mff.cuni.cz}

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