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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 55, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/55\hfil Local smoothing effects]
{A note on local smoothing effects for the
unitary group associated with the KdV equation}

\author[X. Carvajal\hfil EJDE-2008/55\hfilneg]
{Xavier Carvajal}

\address{Xavier Carvajal \newline
 Instituto de Matem\'atica - UFRJ
 Av. Hor\'acio Macedo, Centro de Tecnologia
 Cidade Universit\'aria, Ilha do Fund\~ao,
 Caixa Postal 68530
 21941-972 Rio de Janeiro,  RJ,  Brasil}
\email{carvajal@im.ufrj.br}

\thanks{Submitted March 13, 2008. Published April 17, 2008.}
\subjclass[2000]{35A07, 35Q53}
\keywords{Modified KdV equation; KdV equation; local smoothing effect}

\begin{abstract}
 In this note we show interesting local smoothing effects for the
 unitary group associated to Korteweg-de Vries type equation.
 Our main tools are the Hardy-Littlewood-Sobolev and Hausdorff-Young 
 inequalities.  Using our local smoothing effect and a dual version, 
 we estimate the growth of the norm of solutions of the
 complex modified KdV equation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}


\section{Introduction}

In this note we describe some results on local smoothing effects
for solutions  of the initial value problem (IVP)
\begin{equation}\label{1grupoU}
\begin{gathered}
\partial_t u+b \partial_x^3 u=0,\\
u(x,0)  = u_0(x).
\end{gathered}
\end{equation}
We define the unitary group $U(t)u_0$ as the solution of the linear
initial-value problem \eqref{1grupoU},
in this way
\begin{align}\label{grupoU}
\widehat{U(t)u_0}(\xi)=e^{it(b\xi^3)}\widehat{u_0}(\xi).
\end{align}
Kenig et al. \cite{KPV1} (see also \cite{CL} and \cite{G})
  proved  the following local smoothing effect
\begin{align}\label{x3}
\|\partial_x U(t')u_0\|_{L_x^\infty \mathcal{L}_t^2}
\leq\|\partial_x U(t')u_0\|_{L_x^\infty L_t^2} \le  c
\|u_0\|_{L^2}.
\end{align}
They also proved that
\begin{equation}\label{1x41}
\big\|\partial_x^2 \int_0^t U(t-t')f(t',x)dt'\big\|_{L_x^\infty
L_{t}^2} \leq c \|f\|_{L_x^1 L_{t}^2.}
\end{equation}
In this work we obtain a local smoothing effect (Theorem \ref{1t}),
more general than local smoothing effect \eqref{x3}.
We also consider the IVP for the complex modified Korteweg-de Vries type
equation:
\begin{equation}\label{1.211}
\begin{gathered}
\partial_t u+b \partial_x^3 u+\gamma \partial_x(|u|^2 u)=0,\\
u(x,0)  = u_0(x),
\end{gathered}
\end{equation}
where $u$ is a complex valued function and $b, \gamma $ are real
parameters  with $b \gamma \neq 0$.

Using our local smoothing effect we also proved an interesting
result on growth norms (Theorem \ref{Tprinc}).

The flow associated with  (\ref{1.211})
leads to the  quantity
\begin{equation}
I_1(u) = \int_{\mathbb{R}}|u(x,t)|^2 dx, \label{1.2}
\end{equation}
which is conserved in time. Also, when $b\cdot \gamma \neq 0$ we
have the  time invariant quantity
\begin{equation}
I_2(u) = k_1\int_{\mathbb{R}}|\partial_x u(x,t)|^{2} dx
+k_2\int_{\mathbb{R}}|u(x,t)|^4dx,
\label{1.3}
\end{equation}
where $k_1=3b \gamma$ and $k_2=-3\gamma^2/2$.
The main results in this work are stated as  follows.

\begin{theorem}\label{1t}
Let $U(t)u_0$ be the solution of the linear problem associated
to \eqref{1grupoU} and let $p \ge 2$ and $1/p+1/q=1$.
\\
If $2<p<\infty$ and $4/q-2<s<1/q+1$ then
\[
\|\partial_xU(t')u_0\|_{L_{x}^{\infty} \mathcal{L}_{t}^{p}}\le
c_{p,s} (1+t)^{1/p}\|D^s u_0\|_{L^q}.
\]
If $p=2$ and $0 \le s<3/2$, then
\begin{equation} \label{xav}
\|\partial_xU(t')u_0\|_{L_{x}^{\infty} \mathcal{L}_{t}^{2}}\le
c_s t^{s/3}\|D^s u_0\|_{L^2}.
\end{equation}
If $p=\infty$ and $3/2<s$, then
\begin{equation} \label{xav9}
\|\partial_xU(t')u_0\|_{L_{x}^{\infty} \mathcal{L}_{t}^{\infty}}\le
c_s \|u_0\|_{H^s}.
\end{equation}
\end{theorem}

\begin{theorem}\label{Tprinc}
Let $u\in \mathcal{C}(\mathbb{R}, H^2(\mathbb{R}))$ be solution of
 \eqref{1.211} and $T>0$. Then for all $t \in (0,T)$ there exist
 a function
 $\delta=\delta(\|u\|_{L_x^2\mathcal{L}_T^{\infty}},
  \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}})$ such that
\begin{equation} \label{3eq:1t1}
\|u(t)\|_{\dot{H}^\theta} \le   \|u_0\|_{\dot{H}^\theta}
+\delta t\|u_0\|_{L^2}^3,
\end{equation}
where $0\le \theta \le 1$.
\end{theorem}

The notation  used here is standard in partial differential equations.
We will use the Lebesgue space-time
$L_{x}^{p}\mathcal{L}_{\tau}^{q}$ endowed with the norm
$$
\|f\|_{L_{x}^{p}\mathcal{L}_{\tau}^{q}} =
\big\| \|f\|_{\mathcal{L}_{\tau}^{q}} \big\|_{L_{x}^{p}}
= \Big( \int_{\mathbb{R}} \Big( \int _{0}^{\tau} |f(x,t)|^{q} dt
\Big)^{p/q} dx \Big)^{1/p}.
$$
We will use the notation $\|f\|_{L_{x}^{p}L_{t}^{q}}$ when the
integration in the time variable is on the whole real line.
The notation $\|u\|_{L^p}$ is used when there is no doubt about
the variable of integration.

\section{Smoothing Local Effects}

In this section we prove new smoothing local effects for the unitary group associated with the Korteweg-de Vries equation (Theorem \ref{1t}),
which will be fundamental in the proof of Theorem
\ref{Tprinc}.

\subsection*{Linear Estimates}
The next lemma is a preliminary result to be used in the proof
 of Theorem \ref{Tprinc}.

\begin{lemma}\label{teox3}
Let $u(x,t')=U(t')u_0(x)$ be the solution of \eqref{1grupoU}.
We have the maximal function estimates
\begin{equation}\label{efm}
\| U(t')u_0\|_{L_x^4 L_t^\infty} \leq c \|D^{1/4}u_0\|_{L^{2}},
\end{equation}
and for $s>3/4$ and $\rho>3/4$
\begin{equation}\label{efm1}
\| U(t')u_0\|_{L_x^2 \mathcal{L}_t^\infty} \leq c (1+t)^{\rho}\|u_0\|_{H^{s}}.
\end{equation}
and
\begin{equation}\label{x41}
\big\|\partial_x^2 \int_0^t U(t-t')f(t',x)dt'\big\|_{L_x^\infty
\mathcal{L}_{\tau}^2} \leq c \|f\|_{L_x^1
\mathcal{L}_{\tau}^2}.
\end{equation}
\end{lemma}

\begin{proof}
The proof of  \eqref{efm} and \eqref{efm1} can be found in
\cite{KPV1}. To prove  (\ref{x41}), let $\tau >0$ and
$g(t',\tau,x)=f(t',x)\chi_{[0, \tau]}(t')$. Then
\begin{align*}
\big\|\partial_x^2 \int_0^t U(t-t')f(t',x)dt'\big\|_{L_x^\infty
\mathcal{L}_{\tau}^2}
&=\big\| \Big(\int_{0}^{\tau}\big|\partial_x^2 \int_0^t U(t-t')
g(t',\tau,x)dt'\big|^2 dt\Big)^{1/2}\big\|_{L_x^\infty}\\
&\le  \big\| \Big(\int_{\mathbb{R}}
\big|\partial_x^2 \int_0^t U(t-t')g(t',\tau,x)dt'\big|^2 dt\Big)^{1/2}
\big\|_{L_x^\infty}\\
&=  \big\|\partial_x^2 \int_0^t U(t-t')g(t',\tau,x)dt'\big\|_{L_x^\infty
L_t^2,}
\end{align*}
 and by inequality \eqref{1x41} we obtain (\ref{x41}).
\end{proof}

\begin{proof}[Proof of Theorem \ref{1t}]
Let $\varphi\in \mathcal{C}_0^{\infty}$ with $\varphi(t')=1$ in
$[-t,t]$, $0\le \varphi(t') \le 1$ and
$\mathop{\rm supp}\varphi \subset [-2t,2t]$, then
\[
\|\partial_xU(t')u_0\|_{L_{x}^{\infty}\mathcal{L}_{t}^{p}} \le
\|\varphi(t')\partial_xU(t')u_0\|_{L_{x}^{\infty}L_{t}^{p}}.
\]
Using duality, we consider $g \in L^q$, $\|g\|_{L^q}=1$ and the expression
\begin{align*}
I(x,t):=\big|\int_{\mathbb{R}}g(t')\varphi(t')\partial_xU(t')u_0  dt'
\big|.
\end{align*}
Now using the change of variable $t'=-t'$ we can assume that
\begin{align*}
I(x,t):=\big|\int_{\mathbb{R}}g(t')\varphi(t')\partial_xU(-t')u_0  dt'
\big|.
\end{align*}
Fubinni Theorem and the definition of group $U(t)$, shows that
\begin{equation} \label{inic}
\begin{aligned}
I(x,t)
&=\big|\int_{\mathbb{R}}g(t')\varphi(t')\int_{\mathbb{R}} e^{i x \xi-i\xi^3 t'}i
\xi \widehat{u_0}(\xi) d\xi  dt' \big| \\
&= \big|\int_{\mathbb{R}}e^{i x \xi} \xi \widehat{u_0}(\xi)
\Big(\int_{\mathbb{R}} g(t')\varphi(t') e^{-i\xi^3 t'}dt'\Big) d\xi
\big|\\
&=  \big|\int_{\mathbb{R}}\widehat{u_0}(\xi) \xi e^{i x \xi}
\widehat{\varphi g}(\xi^3)  d\xi\big|,
\end{aligned}
\end{equation}
and by Plancherel's equality,
H\"older inequality and Hausdorff-Young inequality we have
\begin{equation}
\begin{aligned}
I(x,t)
&= \big|\int_{\mathbb{R}}|\xi|^s\widehat{u_0}(\xi)
 \frac{\xi e^{i x \xi}}{|\xi|^s}\widehat{\varphi g}(\xi^3)  d\xi\big| \\
&=\big|\int_{\mathbb{R}}D^s u_0(y) \mathcal{F}
\Big(\frac{ \xi e^{i x \xi}}{|\xi|^s}\widehat{\varphi g}
(\xi^3)\Big)(y)  dy\big| \\
&\le \|D^s u_0\|_{L^q} \big\|\mathcal{F}
\Big(\frac{\xi e^{i x \xi}}{|\xi|^s}\widehat{\varphi g}(\xi^3)\Big)(y)
\big\|_{L^p} \\
&\le  \|D^s u_0\|_{L^q} \big\|\frac{\xi e^{i x
\xi}}{|\xi|^s}\widehat{\varphi g}(\xi^3)\big\|_{L^q}.
\end{aligned}\label{xavier21}
\end{equation}
Now, we make the change of variable $y=\xi^3$ to obtain:
\begin{equation} \label{xaveqt}
\Big\|\frac{\xi e^{i x \xi}}{|\xi|^s}\widehat{\varphi g}(\xi^3)\Big\|_{L^q}^q
= \frac13 \int_{\mathbb{R}}\frac{|\widehat{\varphi g}(y)|^q  dy}{|y|^{\alpha}},
\end{equation}
where $\alpha=(2-(1-s)q)/3$.
Note that if $p=q=2$ and $s=0$, then $\alpha =0$, therefore in this case
\[
I(x,t) \le  c \|u_0\|_{L^2} \|\varphi g\|_{L^2} \le c \|
u_0\|_{L^2} \|g\|_{L^2}=c \|u_0\|_{L^2},
\]
and in this case we obtain \eqref{xav}.

If $p=q=2$ and $0<s<3/2$, then $0<\alpha =2s/3<1$, using properties of
the Fourier transform and the Hardy-Littlewood-Sobolev inequality it
is not hard to deduce the following string of inequalities
\begin{equation} \label{xav7}
\begin{aligned}
\int_{\mathbb{R}}\frac{|\widehat{\varphi g}(y)|^2}{|y|^{2s/3}}  dy
&=  \int_{\mathbb{R}}|\widehat{\varphi g}(y)|^2
\big|\widehat{\frac{1}{|x|^{1-s/3}}}(y)\big|^2 dy \\
&\le \big\|  (\varphi g)*\frac{1}{|x|^{1-s/3}}\big\|_{L^2}^2\\
&\le c_{s}\| \varphi g\|_{L^{6/(3+2s)}}^2 \\
&\le c_{s} \| \varphi \|_{L^{3/s}}^2 \| g \|_{L^{2}}^2 \\
&\le c_{s} t^{2s/3} \| g \|_{L^{2}}^2.
\end{aligned}
\end{equation}
If $p>2$ and  $4/q-2<s<1/q+1$, then $0<\alpha <1$ (observe that
$4/q-2>1-2/q$), we can write the integral in \eqref{xaveqt} as follows
\[
\int_{\mathbb{R}}\frac{|\widehat{\varphi g}(y)|^q  dy}{|y|^{\alpha}}
=  \int_{|y| \le 1}\frac{|\widehat{\varphi g}(y)|^q  dy}{|y|^{\alpha}}+ \int_{|y|> 1}\frac{|\widehat{\varphi g}(y)|^q  dy}{|y|^{\alpha}}\\ \\
:=   I_1^q+ I_2^q,
\]
 hence
\[
I_1^q \le c_{s,q}\|\widehat{\varphi g}\|_{L^{\infty}}^q \le
c_{s,q}\|\varphi g\|_{L^{1}}^q \le
c_{s,q}\|\varphi\|_{L^p}^q\|g\|_{L^q}^q \le c_{s,q}t^{q/p},
\]
note that $s>4/q-2$ implies $\alpha p/(p-q)>1$, therefore using
H\"older inequality and Hausdorf-Young inequality in $I_2^q$ we
obtain
\[
I_2^q \le \|\widehat{\varphi g}\|_{L^p}^q \Big(\int_{|y|>1}
\frac{dy}{|y|^{\alpha p/(p-q)}} \Big)^{1-q/p} \le
c_{s,q}\|\varphi g\|_{L^q}^q \le c_{s,q}\|g\|_{L^q}^q.
\]
If $p=\infty$ and  $s>3/2$, then \eqref{inic} gives
$$
I(x,t) \le \|\widehat{\varphi g}\|_{L^\infty}
\|\widehat{u_0}(\xi) \xi\|_{L^1} \le c_s \| g\|_{L^1}\|u_0\|_{H^s}.
$$
Note that, for $s>1/2$ using immersion we also have
\[
\|\partial_xU(t')u_0\|_{L_{t}^{\infty} L_{x}^{\infty}}\le
c_s \|\partial_xU(t')u_0\|_{H^s} \le
c_s \|u_0\|_{H^{s+1}}.
\]
Hence we have finished the proof of Theorem \ref{1t}.
\end{proof}

\begin{corollary}\label{eq:x53k.1}
Let $0 \le s \le 1$ and $u_0 \in L^2$. Then
\begin{equation}\label{xavtt}
\|D_x^s U(t')u_0\|_{L_x^{\infty}\mathcal{L}_t^2} \le c_s t^{(1-s)/3}
\|u_0\|_{L^2}.
\end{equation}
\end{corollary}

The proof of the above corollary follows from \eqref{xav}.

\begin{corollary}\label{eq:x33k.1}
Let $f \in L_x^1 \mathcal{L}_t^2$ and $U(t')$ be as in \eqref{grupoU}.
Then for $0\le s \le 1$ we have
\begin{equation} \label{4tt}
\big\| D_x^s \int_0^t U(t-t')f(x,t')dt'\big\|_{L_x^2}
\le c_s t^{(1-s)/3} \|f\|_{L_x^1 \mathcal{L}_t^2}.
\end{equation}
\end{corollary}

\begin{proof}
Inequality (\ref{4tt}) follows from (\ref{xavtt}) and a duality
argument. In fact, by Plancherel identity, definition of the group
$U(t)$ and (\ref{xavtt}), we have for $\|g\|_{L^2}=1$:
\begin{align*}
\int_{\mathbb{R}} \Big(D_x^s \int_0^t U(-t')f(x,t')dt'\Big)
\overline{g(x)}dx
&= \int_0^t \int_{\mathbb{R}} f(x,t')\overline{D_x^sU(t')g(x)}dx dt'\\
&\le   \|f\|_{L_x^1 \mathcal{L}_t^2}
 \|D_x^sU(t')g(x)\|_{L_x^{\infty} \mathcal{L}_t^2} \\
&\le  c t^{(1-s)/3}\|f\|_{L_x^1 \mathcal{L}_t^2}  \|g\|_{L^2}.
\end{align*}
\end{proof}


\subsection*{Proof of Theorem \ref{Tprinc}}
The next lemma is used in the  proof.

\begin{lemma}\label{esmax0}
Let $u\in \mathcal{C}(\mathbb{R}, H^2)$ be the solution of \eqref{1.211}.
Then
\begin{equation} \label{esmax22}
\begin{aligned}
\|u\|_{L_x^2\mathcal{L}_t^{\infty}}
&\le   c(1+t)^{3/4+}\|u(0)\|_{H^{3/4+}}
 +c(1+t)^{3/4+}\int_{0}^{t}(\|u(t') \|_{H^{1/2+}}\|u(t') \|_{H^{2}}^2\\
&\quad  +\|u(t') \|_{H^{1/2+}}^2\|u(t') \|_{H^{2}})dt'.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
To prove the first inequality we rely on the integral equation form
\[
u(t)= U(t)u_0- \gamma\int_0^{t}U(t-\tau)\left(\partial_x (|u|^2u)\right)(\tau),
\]
the linear estimate \eqref{efm1} show that if $u(0) \in H^2$ then
for any $t>0$,
\begin{equation} \label{esmax2}
\begin{aligned}
\|u\|_{L_x^2\mathcal{L}_t^{\infty}}
&\le  c(1+t)^{3/4+}\|u(0)\|_{H^{3/4+}}\\
&\quad +c(1+t)^{3/4+}\int_{0}^{t}(\| |u|^2u(t') \|_{L_x^2}
 +\| \partial_x^2(|u|^2u)(t')\|_{L_x^2})dt',
\end{aligned}
\end{equation}
using the immersions $\|u(t)\|_{L_x^{\infty}} \le c\| u(t)\|_{H^{1/2+}}$,
$\|u(t)\|_{L_x^{4}} \le c\| u(t)\|_{\dot{H}^{1/4}}$ it follows that
\begin{align}\label{esmax3}
\| |u|^2u(t') \|_{L_x^2}\le \|u(t') \|_{L_x^{\infty}}\|u^2(t') \|_{L_x^2}
\le c\|u(t') \|_{H^{1/2+}}\|u(t') \|_{L_x^{4}}^2<\infty,
\end{align}
and using Leibniz rule, it is easy to see that
\begin{align*} %\label{esmax4}
\|\partial_x^2(|u|^2u)(t')\|_{L_x^2}
&\le  c\|u u_x^2(t') \|_{L_x^{2}}+c\|u^2u_{xx}(t') \|_{L_x^2}\\
&\le  c\|u(t') \|_{H^{1/2+}}\|u(t') \|_{H^{2}}^2+c\|u(t')
 \|_{H^{1/2+}}^2\|u(t') \|_{H^{2}}< \infty.
\end{align*}
Hence combining this inequality and \eqref{esmax2}, we obtain
 \eqref{esmax22}.
\end{proof}

\begin{lemma}\label{reivax22}
Let $u\in \mathcal{C}(\mathbb{R}, H^2(\mathbb{R}))$ be solution of
 \eqref{1.211} and $0 \le s \le 1$.
Then
\begin{equation} \label{3eq:1t11}
\begin{aligned}
\|D_x^s u(t)\|_{L_x^2}
&\le \|D^s u_0\|_{L^2}\\
&\quad + c  t^{(1-s)/3} \|u\|_{L_x^2\mathcal{L}_t^{\infty}}^2
\Big(\|u_0 \|_{L^2}+t^{1/2}\|u\|_{\mathcal{L}_t^{\infty}\dot{H}^{1/4}}^2
\|u\|_{L_x^2\mathcal{L}_t^{\infty}}\Big).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Without loss of generality we restrict our attention to the real case
$u\in \mathbb{R}$. The equivalent integral equation is
\begin{equation}\label{33k.1}
u(t)=  U(t)u_0- \gamma\int_0^{t}U(t-\tau)
\left(\partial_x (u^3)\right)(\tau)d\tau \\
=:  U(t)u_0+z(t).
\end{equation}
Let $\Gamma(t)=\|u\|_{L_x^2\mathcal{L}_t^{\infty}}$.
From (\ref{33k.1}), Corollary \ref{eq:x33k.1} and H\"older
inequality, we have
\begin{equation}
\begin{aligned}
\|D_x^s u(t)\|_{L_x^2}
&\le  \|D_x^s U(t)u_0\|_{L_x^2}+\|D_x^s z(t)\|_{L_x^2} \\
&\le  \|D^s u_0\|_{L^2}+ c t^{(1-s)/3}\|u^2 u_x\|_{L_x^1\mathcal{L}_t^2} \\
&\le  \|D^s u_0\|_{L^2}+ c t^{(1-s)/3}\Gamma(t)^2
\|u_x\|_{L_x^{\infty}\mathcal{L}_t^2}.
\end{aligned} \label{1eq:t1}
\end{equation}
Using (\ref{x3}), (\ref{x41}) and H\"older inequality, we obtain
\begin{equation}  \label{2eq:t1}
\begin{aligned}
\|\partial_x u\|_{L_x^{\infty}\mathcal{L}_t^2}
&\le  \|\partial_x U(t')u_0\|_{L_x^{\infty}
\mathcal{L}_t^2}+\|\partial_x z\|_{L_x^{\infty}\mathcal{L}_t^2} \\
&\le  c\|u_0\|_{L^2}+ c\|u^3\|_{L_x^1 \mathcal{L}_{t}^2} \\
&\le   c\|u_0\|_{L^2}+ c \|u\|_{L_x^4\mathcal{L}_t^4}^2\Gamma(t)\\
&\le  c\|u_0\|_{L^2}+ c  t^{1/2}\|u\|_{\mathcal{L}_t^\infty L_x^4}^2\Gamma(t)\\
&\le   c\|u_0\|_{L^2}+ c t^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2} \Gamma(t),
\end{aligned}
\end{equation}
where in the last inequality we use immersion
$\|u\|_{L_x^4} \le \|u\|_{\dot{H}^{1/4}}$.
As a consequence of (\ref{1eq:t1}) and (\ref{2eq:t1}) we have
\eqref{3eq:1t11}.
Thus the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Tprinc}]
Let $T>0$. Then there is a $\delta_0=\delta_0(T)>0$ such that
\begin{equation}\label{reivax11}
\|u\|_{L_x^2L^{\infty}([\tau_{1}, \tau_{2}])}<2\|u_0\|_{L^2},\quad
\textrm{for all }  \tau_1,\tau_2 \in[0,T],  \;  |\tau_1-\tau_2|\le \delta_0.
\end{equation}
To verify this we use contradiction, we suppose that for all $n$
there exist $\tau_1^n, \tau_2^n \in [0,T]$, $|\tau_1^n- \tau_2^n|<1/n$ and
\begin{equation}\label{reivax10}
\|u\|_{L_x^2L^{\infty}([\tau_{1}^n, \tau_{2}^n])}\ge 2\|u_0\|_{L^2}.
\end{equation}
Since $(\tau_1^n)$ and $(\tau_2^n)$ are bounded sequences, we can suppose
that there exist a $\tau\in[0,T]$ such that
$\lim_{n\to \infty} \tau_1^n=\lim_{n\to \infty} \tau_2^n=\tau$,
using Lemma \ref{esmax0} and Lebesgue's Dominated Convergence Theorem,
we have that
$$
\|u\|_{L_x^2L^{\infty}([\tau_{1}^n, \tau_{2}^n])}
\to \|u(\tau)\|_{L^2}=\|u_0\|_{L^2} \quad \text{as $n\to \infty$};
$$
however, this contradicts the relation \eqref{reivax10}.

Let $0 \le t_k \le t$ be a sequence with $t_0=0$, $t_{k+1}-t_{k}=\delta_0$
and let $n \approx t/\delta_0$ such that $t_n \le t<t_{n+1}$.
By Lemma \ref{reivax22} and \eqref{reivax11}, it follows that
\begin{align*}
\|D_x^s u(t_k)\|_{L_x^2}
&\le  \|D_x^s u(t_{k-1})\|_{L^2}+ c \delta_0^{(1-s)/3}
\|u\|_{L_x^2L^{\infty}([t_{k-1}, t_{k}])}^2\|u_0 \|_{L^2}\\
&\quad +  \delta_0^{(1-s)/3+1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2}\|u\|_{L_x^2L^{\infty}([t_{k-1},
 t_{k}])}^3\\
&\le    \|D_x^s u(t_{k-1})\|_{L^2}+ c \delta_0^{(1-s)/3} \|u_0\|_{L^2}^3
(1+\delta_0^{1/2}  \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2}),
\end{align*}
similarly we have
\begin{align}\label{reivax12}
\|D_x^s u(t)\|_{L_x^2} \le    \|D_x^s u(t_{n})\|_{L^2}
+ c \delta_0^{(1-s)/3} \|u_0\|_{L^2}^3(1+\delta_0^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2});
\end{align}
therefore,
\begin{align*}
\|D_x^s u(t_n)\|_{L_x^2} - \|D^s u(0)\|_{L_x^2}
&= \sum_{k=1}^{n}\left(\|D_x^s u(t_k)\|_{L_x^2} - \|D_x^s u(t_{k-1})\|_{L^2}\right)\\
&\le   \sum_{k=1}^{n} c \delta_0^{(1-s)/3} \|u_0\|_{L^2}^3(1+\delta_0^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2})\\
&\le  c  t\|u_0\|_{L^2}^3\frac{(1+\delta_0^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2})}{\delta_0^{(2+s)/3}},
\end{align*}
so that we conclude
\begin{align}\label{reivax33}
\|D_x^s u(t_n)\|_{L_x^2} \le \|D^s u(0)\|_{L^2}+
c  t\|u_0\|_{L^2}^3\frac{(1+\delta_0^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2})}{\delta_0^{(2+s)/3}},
\end{align}
combining \eqref{reivax12} and \eqref{reivax33} we obtain
\[
\|D_x^s u(t)\|_{L_x^2} \le \|D^s u(0)\|_{L^2}+  \|u_0\|_{L^2}^3
\frac{c  (t+\delta_0)}{\delta_0^{(2+s)/3}} (1+\delta_0^{1/2}
 \|u\|_{\mathcal{L}_T^\infty\dot{H}^{1/4}}^{2}).
\]
This completes the proof.
\end{proof}

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\end{document}