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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 95, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/95\hfil An extension of the Lax-Milgram theorem]
{An extension of the Lax-Milgram theorem and its
application to fractional differential equations}

\author[N. Nyamoradi, M. R. Hamidi \hfil EJDE-2015/95\hfilneg]
{Nemat Nyamoradi, Mohammad Rassol Hamidi}

\address{Nemat Nyamoradi \newline
Department of Mathematics, Faculty of Sciences,
Razi University, 67149 Kermanshah, Iran}
\email{nyamoradi@razi.ac.ir, neamat80@yahoo.com}

\address{Mohammad Rassol Hamidi  \newline
Department of Mathematics, Faculty of Sciences,
Razi University, 67149 Kermanshah, Iran}
\email{mohammadrassol.hamidi@yahoo.com}

\thanks{Submitted February 1, 2015. Published April 13, 2015.}
\makeatletter
\@namedef{subjclassname@2010}{\textup{2010} Mathematics Subject Classification}
\makeatother
\subjclass[2010]{34A08, 35A15, 35B38}
\keywords{Lax-Milgram theorem; fractional differential equation}

\begin{abstract}
 In this article, using an iterative technique, we introduce an
 extension of the Lax-Milgram theorem which can be used for proving
 the existence  of solutions to boundary-value problems.
 Also, we apply of the obtained result to the
 fractional differential equation
 \begin{gather*}
 {}_t D_T^{\alpha}{}_0 D_t^{\alpha}u(t)+u(t)
 =\lambda f (t, u(t)) \quad t \in (0,T),\\
 u(0)=u(T)=0,
 \end{gather*}
 where ${}_tD_T^\alpha$ and ${}_0D_t^\alpha$ are the right and
 left Riemann-Liouville fractional derivative of order
 $\frac{1}{2}< \alpha \leq 1$ respectively, $\lambda$ is a
 parameter and $f:[0,T]\times\mathbb{R}\to\mathbb{R}$
 is a continuous function. Applying a regularity argument to this
 equation, we show that every weak solution is a classical solution.
\end{abstract}

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


 \section{Introduction}

Fractional differential equations form a very important and
significant part of mathematical analysis and its applications
to real-world problems. On the other hand, the Lax-Milgram theorem
is a very useful tool in the wide area of functional analysis such
as the theory of operator equations in Banach spaces. It is also
used in the studies of fractional differential equations, ordinary
and partial differential equations (see \cite{B, ER, E} and the references therein).
For example, Ervin and Roop \cite{ER} using the
Lax-Milgram theorem,  investigated the existence of
solutions to the following fractional boundary value problem
\begin{gather*}
-Da\big(p_0 D_t^{-\beta}+q_t D_1^{-\beta}\big)Du(t)+b(t)Du(t)+c(t)u(t)=f (t),\\
u(0)=u(1)=0,
\end{gather*}
where $D$ represents a single spatial derivative, ${}_0D_t^{-\beta}$ 
and ${}_tD_1^{-\beta}$ are the left and right Riemann-Liouville 
fractional integrals of order $0\leq \beta <1$, respectively,
 $f,c \in C([0,1])$ and $b \in C^1([0,1])$, $a>0$ and $0 \leq p,q \leq 1$ 
with $p+q=1$.

Recently, Jiao and Zhou \cite{JZ}, for the first time, showed that
the critical point theory is an effective approach for studying
the existence for the following fractional boundary value problem
\begin{gather*}
{}_tD_T^{\alpha} \big({}_0 D_t^\alpha u (t) \big)= \nabla F (t, u (t)),\quad 
t \in (0, T),\\
u (0) = u (T) = 0,
\end{gather*}
and obtained the existence of at least one nontrivial solution.

 We know that, we can only use the Lax-Milgram theorem to prove the 
existence of solutions to equations in the form
$$
Lu=f (t),
$$
where $f$ is independent of $u$ and $L$ is an operator. In our
investigations, we apply the iterative technique to generalize the
Lax-Milgram theorem \cite{D}. Moreover, we are going to study the
solvability of the following fractional differential equation
\begin{equation}\label{ip0}
    \begin{gathered}
{}_t D_T^{\alpha}{}_0 D_t^{\alpha}u(t)+u(t)=\lambda f (t, u(t)) \quad t \in (0,T),\\
u(0)=u(T)=0,
    \end{gathered}
\end{equation}
where ${}_tD_T^\alpha$ and ${}_0D_t^\alpha$ are the right and left 
Riemann-Liouville fractional derivative of order
 $1/2< \alpha \leq 1$ respectively, $\lambda$ is a parameter and 
$f: [0,T]\times\mathbb{R}\to\mathbb{R}$ is a  continuous function.

This article is organized as follows:
Section 2 is devoted to our main results.  
In Section 3, we prove any weak solution of Problem \eqref{ip0}  
is a classical solution.

\section{Main results}

 In this section, firstly, we recall some notation
and theorems to obtain the results of this work. Let 
$(X, \|\cdot\|_{X})$ be a real Banach space with dual space $X^*$.
Denote by $B_r(x_0)$ the ball  $B_r(x_0) = \{x \in X : \|x -x_0 \|_{X}\leq r \}$.

In the following, we state the Lax-Milgram theorem.

\begin{theorem}[{\cite[Proposition 1.2.41]{D}}]
 Let $H$ be a complex Hilbert space and let $B:H\times H\to\mathbb{C}$ 
be a mapping with the following properties:
\begin{itemize}
\item[(i)] The mapping $x\mapsto B(x, y)$ is linear for any $y\in H$.

\item[(ii)] $B(x, \alpha_1 y_1+ \alpha_2 y_2)= \bar{\alpha}_1B(x,
y_1)+\bar{\alpha}_2B(x, y_2)$ for every $x, y_1,y_2\in H$,
$\alpha_1, \alpha_2\in \mathbb{C}$.

\item[(iii)] There is a constant $c$ such that $|B(x, y)|\leq c\|x\| \; \|y\|$ 
for every $x,y\in H$.
\end{itemize}
Then there is $A\in L(H)$, $\|A\|_{L(H)}\leq c$, such that
$$
B(x, y)=(x,Ay),\quad x,\;y\in H.
$$
Moreover, if
\begin{itemize}
\item[(iv)]  there is a positive constant $d$ such that
$$
B(x, x)\geq d\|x\|^2\quad \forall x\in H,
$$
\end{itemize}
then $A$ is invertible,
$A^{-1}\in L(H)$ and $\|A^{-1}\|_{L(H)}\leq \frac{1}{d}$.
\end{theorem}

The main result of this section reads as follows.

\begin{theorem}\label{mt1}
Suppose that $H$ is a Hilbert space, $B(u,v)$ is a continuous coercive 
bilinear form on $H$ and $F:H\to H^*$ satisfying the following conditions:
\begin{itemize}
\item[(F1)] There exists a constant $N>0$ such that
$$
\|F(u)\|_{H^*}\leq N \quad \forall u \in B_1(0),
$$
where $B_1(0) = \{u \in H : \|u\|_H\leq 1 \}$

\item[(F2)] If  $\{u_k\}$ is a sequence in $H$ such that 
$u_k\rightharpoonup u$ weakly in $H$, then the sequence 
$\{F(u_k)\}$ has a subsequence $\{F(u_{k_n})\}$ such that 
$F(u_{k_n}) \rightharpoonup F(u)$ weakly in $H^*$.
\end{itemize}
Then, there exists a constant $L>0$ such that for any $\lambda \in \mathbb{R}$ 
with $|\lambda|\leq L$, there exists an element $\tilde{u}\in H$ such that
$$
B(\tilde{u},v)=\lambda\langle F(\tilde{u}),v\rangle \quad \forall v\in H.
$$
\end{theorem}

\begin{proof}
Take any $u_0 \in H$ with  $\|u_0\|_H\leq 1$. From the Riesz
representation theorem, there exists a unique element $G(u_0)\in H$ such that 
$\|G(u_0)\|_H=\|F(u_0)\|_{H^*}$ and
\begin{equation}\label{me3}
\langle F(u_0),v\rangle =(G(u_0),v)\quad\forall  v\in H,
\end{equation}
where $(\cdot,\cdot)$ denotes the inner product of $H$.

 By the hypotheses of the theorem, there are two constants $a>0$ and $b>0$ 
such that for all $u,v\in H$, we have
\begin{gather*}
|B(u,v)|\leq a\|u\|_H\|v\|_H,\\
|B(u,u)|\geq b\|u\|^2_H.
 \end{gather*}
Thus, the Lax-Milgram theorem (see \cite{D})  yields the existence of a 
continuous and invertible linear operator $A$ on $H$ such that 
$\|A\|_{L(H)}\leq a$, $\|A^{-1}\|_{L(H)}\leq 1/b$ and
\begin{equation}\label{me4}
B(u,v)=(Au,v)\quad \forall u,v \in H.
\end{equation}
Then, one can conclude the existence of a unique element $u_1\in H$ such 
that $Au_1=\lambda G(u_0)$. 
So, in view of \eqref{me3} and \eqref{me4}, we have that
$$
B(u_1,v)=(Au_1,v)=\lambda (G(u_0),v)
=\lambda \langle F(u_0),v\rangle \forall v\in H
$$
such that
$$
\|u_1\|_H=|\lambda|\;\|A^{-1}G(u_0)\|_H\leq \frac{N|\lambda|}{b}.
$$

Set $L=b/N$. Hence if $|\lambda| \leq L$, one can get
\begin{gather*}
B(u_1,v)=\lambda \langle F(u_0),v\rangle \quad \forall v\in H,\\
\|u_1\|_H\leq 1.
\end{gather*}
Similarly, there exists an element $u_2\in H$ such that
\begin{gather*}
B(u_2,v)=\lambda \langle F(u_1),v \rangle\quad \forall v\in H,\\
\|u_2\|_H\leq 1.
\end{gather*}
So by induction, we have a sequence $\{u_n\}$ such that
\begin{equation}\label{me5}
    \begin{gathered}
B(u_n,v)=\lambda \langle F(u_{n-1}),v\rangle \quad \forall v\in H,\\
\|u_n\|_H\leq 1.
   \end{gathered}
\end{equation}
The reflexivity of $H$ implies that there exists a subsequence of
$\{u_n\}$ still denoted by $\{u_n\}$  such that
$u_n\rightharpoonup \tilde{u}$ weakly in $H$. Finally, in view of
(F2), the desired conclusion follows from \eqref{me5} and
letting $n \to +\infty$. 
\end{proof}

Now, by using Theorem \ref{mt1}, we  prove the existence of
one solution to Problem \eqref{ip0}. To this end, we need the
following preliminaries.

\begin{definition}[\cite{Podlubny, 17}]\label{pd1} \rm
%(Left and Right Riemann-Liouville Fractional Integrals ).
Let $\phi$ be a function defined on $[0,T]$. 
Then, the left and right Riemann-Liouville fractional integrals of order $0< \alpha < 1$ on the interval $[0,T]$ are respectively defined by
\begin{gather*}
{}_0I_t^{\alpha} \phi (t)  
=  \frac{1}{\Gamma (\alpha)} \int_0^t (t - \xi)^{\alpha - 1} \phi (\xi) d \xi, \\
{}_t I_{T}^{\alpha} \phi (t)  
=  \frac{1}{\Gamma (\alpha)} \int_t^{T} (\xi - t)^{\alpha - 1} \phi (\xi) d \xi.
\end{gather*}
The left and right Riemann-Liouville fractional derivatives of order 
$0 < \alpha < 1$ on the interval $[0,T]$ are respectively defined by
\begin{gather*}
{}_0 D_t^{\alpha} \phi (t)  
=  \frac{d}{dt}\Big({}_0I_t^{1 - \alpha} \phi (t)\Big), \\
{}_t D_{T}^{\alpha} \phi (t) 
=  - \frac{d}{dt}\Big({}_t I_{T}^{1 - \alpha} \phi (t)\Big).
\end{gather*}
\end{definition}

Taking $p=2$ in Definition 3.1, \cite[Propositions 3.1 and 3.3]{ZJ}, 
 we deduce the following definition and theorems.

\begin{definition}[\cite{ZJ}] \label{pd2}\rm
Let $0<\alpha \leq 1$. The fractional derivative space $E^\alpha_0$ is 
defined by the closure of $C^\infty_0([0,T])$ with respect to the norm
\[
\|u\|_{E^\alpha_0} = \Big(\|u\|_{L^2(0,T)}^2 
+ \|{}_0 D_t^{\alpha}u\|_{L^2(0,T)}^2 \Big)^{1/2}.
\]
\end{definition}


\begin{theorem}[\cite{ZJ}]\label{0t1}
Let $0<\alpha \leq 1$. The fractional derivative space $E^\alpha_0$ 
is a reflexive and separable Banach space.
\end{theorem}

\begin{remark}\label{pr1}\rm
In fact, the space $E^\alpha_0$ is a separable Hilbert space with the inner product
\[
 (u, v )_{E^\alpha_0} = \int_0^T \Big({}_0 D_t^\alpha u(t)\;
 {}_0 D_t^\alpha v(t)+u(t) v(t) \Big) dt.
\]
\end{remark}

\begin{theorem}[\cite{ZJ}]\label{pt2}
Assume that $\alpha > \frac{1}{2}$ and the sequence $\{u_k\}$ 
converges weakly to u in $E^{\alpha}_0$, then $u_k\to u$ in $C([0, T ])$.
\end{theorem}

\begin{remark}[\cite{ZJ}] \label{pr2} \rm
We have
$$
\|u\|_\infty\leq \frac{T^{\alpha-{\frac{1}{2}}}}{\Gamma (\alpha)(2\alpha-1)^{1/2}} 
\|u\|_{E^\alpha_0}\quad \forall u\in E^{\alpha}_0.
$$
\end{remark}

\begin{definition}\label{md1} \rm
A function $u\in E_0^\alpha$ is a weak solution of 
Problem \eqref{ip0}, provided that
\begin{equation}\label{me7}
\int_0^T\Big({}_0 D_t^{\alpha} u(t)_0 D_t^{\alpha}v(t)+u(t)v(t)\Big)dt
=\lambda \int_0^T f(t,u(t))v(t)dt.
\end{equation}
 for any $v\in E_0^\alpha$.
\end{definition}

Set
$$
\Delta =\max\Big\{f(t,s):(t,s)\in[0,T]
\times\Big[{\frac{-T^{\alpha-\frac{1}{2}}}
{\Gamma (\alpha)(2\alpha-1)^{1/2}},
\frac{T^{\alpha-\frac{1}{2}}}{\Gamma (\alpha)(2\alpha-1)^{1/2}}}\Big]\Big\}.
$$

\begin{theorem}\label{mt2}
Suppose that $\frac{1}{2}<\alpha \leq 1$ and 
$f \in C([0,T]\times \mathbb{R},\mathbb{R})$, then for any  
$|\lambda|\leq\frac{1}{\Delta T^{1/2}}$,
Problem \eqref{ip0} has at least one weak solution.
\end{theorem}

\begin{proof}
First, we define 
$B(u,v)= \int_0^T\Big({}_0 D_t^{\alpha} u(t)_0 D_t^{\alpha}v(t)+u(t)v(t)\Big)dt$.  
Since,
\begin{gather*}
|B(u,v)|\leq \|{}_0 D_t^{\alpha}u\|_{2}\|{}_0 D_t^{\alpha}v\|_{2}+\|u\|_{2}\|v\|_{2}\leq 2\|u\|^2_{E_0^\alpha}\|v\|^2_{E_0^\alpha},\\
|B(u,u)|\geq \|u\|^2_{E_0^\alpha},
\end{gather*}
it follows that $B$ is a continuous coercive bilinear form on $E_0^\alpha$ 
with $a=2$ and $b=1$.

We now define
\begin{gather*}
F:E_0^\alpha \to (E_0^\alpha)^*,\\
\langle F(u),v\rangle =\int_0^T f(t,u(t))v(t)dt.
\end{gather*}
Assume $u\in E_0^\alpha$ with $\|u\|_{E^{\alpha}_0}\leq 1$. 
Then, in view of Remark \ref{pr2}, one has
\begin{equation}\label{G}
\|u\|_{\infty}  \leq \frac{T^{\alpha-\frac{1}{2}}}{\Gamma
(\alpha)(2\alpha-1)^{1/2}}\|u\|_{E^{\alpha}_0}\leq
\frac{T^{\alpha-\frac{1}{2}}}{\Gamma
(\alpha)(2\alpha-1)^{1/2}}.
\end{equation}
 and we have $|u(t)|\leq \frac{T^{\alpha-\frac{1}{2}}}
{\Gamma (\alpha)(2\alpha-1)^{1/2}}$  for any $t\in [0,T]$.
 So, we can conclude that $|f(t,u(t))|\leq {\Delta} $ for any $t\in [0,T]$.

For any $v \in E^{\alpha}_0 $ with $\|v\|_{E^{\alpha}_0}=1$, by
the H\"{o}lder inequality, we have
$$
|\langle F(u),v\rangle |
=\Big|\int^T_0f(t,u(t))v(t)dt\Big|
\leq \Big(\int^T_0|f(t,u)|^2dt\Big)^{1/2}\|v\|_{L^2(0,T)}
\leq \Delta T^{1/2}.
$$
Taking $N= \Delta T^{1/2}$, Condition (F1) holds.

Suppose $\{u_k\}$ is a sequence in $E^{\alpha}_0$ such that 
$u_k\rightharpoonup u$ weakly in $E^{\alpha}_0$. Then, 
Theorem $\ref{pt2}$ yields that  for any $t \in[0,T]$
$$
u_k(t)\to u(t)\quad\forall  t\in [0,T].
$$
By using it and that $f$ is continuous, we have
\begin{equation}\label{me2}
    f(t,u_k(t))\to f(t,u(t))\quad\textrm{as }k\to \infty, \quad\forall t\in [0,T].
\end{equation}

On the other hand, $\{u_k\}$  is a bounded subset of $E^{\alpha}_0$ 
(see \cite[Theorem 3.18]{R}). In other words, there exists a constant 
$K>0$ such that $\|u_k\|_{E_0^\alpha } \leq K$ for any $k\in \mathbb{N}$. 
From \eqref{G}, we have 
$\|u_k\|_\infty \leq \frac{KT^{\alpha-\frac{1}{2}}}{\Gamma (\alpha)(2\alpha-1)^{1/2}}$
for any $k\in \mathbb{N}$. Therefore, we have that there exists a constant
 $\Delta_0>0$ such that
 \begin{equation}\label{me1}
 |f(t,u_k(t))|<\Delta_0 \quad \text{a.e. on } [0,T],\; k=1,2,3,\dots .
 \end{equation}


From \eqref{me1}, \eqref{me2} and the Lebesgue's dominated theorem, 
we  conclude that
$$
\int_0^T\Big|f(t,u_k)-f(t,u)\Big|^2dt\to0.
$$
For any $v \in E^{\alpha}_0$ with  $\|v\|_{E^{\alpha}_0}=1$, we have
$$
|\langle F(u_k)-F(u),v\rangle|
=\Big|\int^T_0\Big(f(t,u_k(t))-f(t,u(t)\Big)v(t)dt\Big|
\leq \|f(t,u_k)-f(t,u)\|_2\to0,
$$
which yields that $F$ satisfies (F2).
Then by Theorem \ref{mt1}, we obtain the desired conclusion.
\end{proof}

 \section{Regularity}
 The main result of this section reads as follows.

\begin{theorem}\label{rt1}
Under the assumptions of Theorem \ref{mt2}, every weak solution of Problem 
\eqref{ip0} is a classical solution.
\end{theorem}

To prove the above theorem, we need the following 
lemmas and definitions.

\begin{definition}[\cite{IW}] \label{rd1} \rm
Let $u\in L^2(0,T)$, $v, w\in L^2(0,T)$ and
\begin{gather*}
\int_0^T u(t)_tD_{T}^\alpha \varphi(t)dt
=\int_0^T v(t) \varphi(t)dt\quad \forall  \varphi  \in C_0^\infty(0,T),\\
\int_0^T u(t)_0D_t^\alpha \varphi(t)dt
=\int_0^T w(t) \varphi(t)dt\quad \forall  \varphi  \in C_0^\infty(0,T).
\end{gather*}
The functions $v$ and $w$ given above will be called the weak left and 
the weak right fractional derivative of order $\alpha \in (0,1]$ of $u$ 
respectively. Here, we denote them by ${}_0\overline{D}_t^\alpha u(t)$ 
and ${}_t\overline{D}_{T}^\alpha u(t)$ respectively.
\end{definition}

In view of Definition \ref{pd2}, $u\in E_0^\alpha$ means that $u$ is the limit
 of a Cauchy sequence $\{u_n\} \subset C_0^\infty (0,T)$. In other words, 
$u_n\to u$ in $L^2(0,T)$ and there exists an element $w\in L^2(0,T)$ 
such that ${}_0D_t^\alpha u_n\to w$ in $L^2(0,T)$.
Then for any $\varphi \in C_0^\infty(0,T)$, we have
\begin{align*}
\int_0^Tw(t)\varphi(t)dt
&=\lim_{n \to \infty}\int_0^T{}_0D_t^\alpha u_n(t)\varphi(t)dt\\
&=\lim_{n \to \infty}\int_0^Tu_n(t)_tD_{T}^\alpha \varphi(t)dt\\
&=\int_0^Tu(t)_tD_{T}^\alpha\varphi(t)dt.
\end{align*}
So, $w={}_0\overline{D}_t^\alpha u$ however it is not clear whether 
${}_0D_t^\alpha u(t)$ exists in the usual sense, for any 
$t\in[0,T]$ or not (see \cite[p. 202]{B} for the case $\alpha =1$).


\begin{remark}[{\cite[Lemma 2.7]{ER}}] \label{rr1} \rm
Let $u\in E_0^\alpha $, then for any $v\in E_0^\alpha $,  we have
\begin{align*}
\int_0^T u(t)_tD_{T}^\alpha v(t)dt 
&= \int_0^T{}_0I_t^\alpha {}_0D_t^\alpha u(t)_tD_{T}^\alpha v(t)dt\\
&= \int_0^T{}_0D_t^\alpha u(t)_tI_{T}^\alpha {}_tD_{T}^\alpha v(t)dt\\
&= \int_0^T{}_0D_t^\alpha u(t)\; v(t)dt.
\end{align*}
Since $C_0^\infty(0,T) \subset E_0^\alpha$, 
we conclude ${}_0D_t^\alpha u(t)={}_0\overline{D}_t^\alpha u(t)$ a.e. on $[0,T]$.
\end{remark}

\begin{lemma}\label{rl1}
Let $u\in E_0^\alpha$, then ${}_0\overline{D}_t^\alpha u$ is almost everywhere 
equal to the weak derivative of ${}_0I_t^{1-\alpha}u$ in the $H^1(0,T)$ 
sense. In other words
$$
{}_0\overline{D}_t^\alpha u(t)=\overline{D}({}_0I_t^{1-\alpha}u(t))
\quad\text{a.e. on } [0,T].
$$
\end{lemma}

\begin{proof}
For any $\varphi \in C_0^\infty(0,T)$, we have 
(see \cite[Remark 3.1]{ZJ} and \cite[Theorem 2.1]{17})
\begin{align*}
\int_0^T{}_0\overline{D}_t^\alpha u(t) \varphi(t)dt
&= \int_0^Tu(t)_tD_{T}^\alpha\varphi(t)dt\\
&= \int_0^Tu(t)_t^C D_{T}^\alpha\varphi(t)dt \\
&= -\int_0^Tu(t)_tI_{T}^{1-\alpha}(D\varphi(t))dt\\
&= -\int_0^T{}_0I_t^{1-\alpha}u(t) D\varphi(t)dt.
\end{align*}
\end{proof}

\begin{lemma}\label{rl2}
Let $0<\alpha \leq 1$ .
\begin{itemize}
\item[(i)] If $\frac{1}{2}<\alpha \leq 1$ and $u \in L^2(0,T)$, 
then ${}_0I_0^{\alpha}u(0)=0$.

\item[(ii)] If $u\in C([0,T])$, then ${}_0I_t^{\alpha}u\in C([0,T])$.

\item[(iii)] If $u\in C^1([0,T])$ and $u(0)=0$, then
 ${}_0I_t^{\alpha}u\in C^1([0,T])$.
\end{itemize}
\end{lemma}

\begin{proof}
It is easy to see that
$$
|{}_0I_t^{\alpha}u(t)|
\leq \frac{t^{\alpha-\frac{1}{2}}}{\Gamma (\alpha) (2\alpha-1)^{1/2}}\|u\|_2,
$$
which completes the proof of (i).

Let $t_0\in [0,T]$ and   $\{t_n\}$ be a sequence in  $[0,T]$ such that 
$t_n\to t_0$. We take
 $$
M=\max_{0\leq t \leq T}|u(t)|.
$$
 From $u(t_n-s)\to u(t_0-s)$ a.e. on $[0,T]$, 
\[
s^{\alpha-1}|u(t_n-s)-u(t_0-s)|\leq 2Ms^{\alpha-1},
\] 
and the Lebesgue's dominated theorem, we  conclude that
\begin{align*}
&\Big|\int_0^{t_n}s^{\alpha-1}u(t_n-s)ds
-\int_0^{t_0}s^{\alpha-1}u(t_0-s)ds\Big|\\
&\leq \int_0^{t_n}s^{\alpha-1}|u(t_n-s)-u(t_0-s)|ds
 + \int_{t_0}^{t_n}|s^{\alpha-1}u(t_0-s)|ds\\
&\leq \int_0^{\xi}s^{\alpha -1}|u(t_n-s)-u(t_0-s)|ds
+\int_{t_0}^{t_n}|s^{\alpha-1}u(t_0-s)|ds
\to 0,
\end{align*}
where $\xi=\max\{t_0, t_1, t_2, \ldots\}$. This concludes the
proof of (ii).

Suppose that
$$
K=\max_{0\leq t \leq T}|u'(t)|,
$$
then  $|\frac{u(t_n-s)-u(t_0-s)}{t_n-t_0}|\leq K$. Hence by the Lebesgue's 
dominated theorem, we  conclude that
\begin{align*}
&\Gamma(\alpha)\Big(\frac{{}_0I_{t_n}^{\alpha}u(t_n)-{}_0I_{t_0}^{\alpha}
u(t_0)}{t_n-t_0}\Big)\\
&=\int_0^{t_0}s^{\alpha-1}\frac{u(t_n-s)
-u(t_0-s)}{t_n-t_0}ds
+\int_{t_0}^{t_n}s^{\alpha-1}\frac{u(t_n-s)-u(t_0-s)}{t_n-t_0}ds \\
&\quad +\int_{t_0}^{t_n}s^{\alpha-1}\frac{u(t_0-s)-u(0)}{t_0-s}
 \frac{t_0-s}{t_n-t_0}ds \to \int_0^{t_0}s^{\alpha-1}u'(t_0-s)ds.
\end{align*}
Thus $\Big({}_0I_t^\alpha u(t)\Big)'(t_0)={}_0I_{t_0}^\alpha u'(t_0)$. 
From this and part (ii), we can conclude (iii).
\end{proof}

\begin{lemma}\label{rl3}
Suppose that for some $u\in L^2(0,T)$, ${}_0\overline{D}_t^\alpha u$
exists and is almost everywhere equal to a function in $C([0,T])$.
Then:
\begin{itemize}
\item[(i)] $u$  is almost everywhere equal to a function  $\Tilde{u}\in C([0,T])$.

\item[(ii)] ${}_0D_t^\alpha {u}(t)$ exists for any $t\in [0,T]$ and 
${}_0D_t^\alpha {u}\in C([0,T])$.
\end{itemize}
\end{lemma}

\begin{proof} Lemma \ref{rl1} implies that
$\overline{D}({}_0I_t^{1-\alpha}u)$ is almost everywhere equal
to a function in $C([0,T])$. Therefore, ${}_0I_t^{1-\alpha}u$
is almost everywhere equal to a function in $C^1([0,T])$ (see
\cite[p. 204]{B}). Thus by Lemma \ref{rl2},
$\int_0^tu(s)ds={}_0I_t^{\alpha}{}_0I_t^{1-\alpha}u(t)\in
C^1([0,T])$. Take $\Tilde{u}(t)=D(\int_0^tu(s)ds)$  (see
\cite[Lemma 8.2]{B}), this completes the proof of (i).

Lemma \ref{rl2} implies that ${}_0I_t^{1-\alpha}\Tilde{u}(t)\in C([0,T])$. Since
\begin{equation}\label{666}
{}_0I_t^{1-\alpha}u(t)={}_0I_t^{1-\alpha}\Tilde{u}(t)\quad \forall  t\in [0,T],
\end{equation}
we can conclude ${}_0I_t^{1-\alpha}{u}(t)\in C([0,T])$. 
By using it and the fact that ${}_0I_t^{1-\alpha}{u}$ is almost everywhere
 equal to a function in $C^1([0,T])$, we can  conclude 
${}_0I_t^{1-\alpha}{u}(t)\in C^1([0,T])$. The desired conclusion 
can be obtained from ${}_0D_t^{\alpha}{u}(t)=D\big({}_0I_t^{1-\alpha}{u}(t)\big)$.
\end{proof}

Quite similar to Lemma \ref{rl2} and Lemma \ref{rl3}, we have the
following lemmas.

\begin{lemma}\label{rl4}
Let $1/2<\alpha \leq 1$.
\begin{itemize}
\item[(i)] If $u \in L^2(0,T)$, then ${}_{T}I_{T}^{\alpha}u(T)=0$.


\item[(ii)] If $u\in C([0,T])$, then ${}_tI_{T}^{\alpha}u\in C([0,T])$.

\item[(iii)] If $u\in C^1([0,T])$ and $u(T)=0$, then 
${}_tI_{T}^{\alpha}u\in C^1([0,T])$.
\end{itemize}
\end{lemma}

\begin{lemma}\label{rl5}
Suppose that $u\in L^2(0,T)$, ${}_tD_{T}^\alpha u$ exists and is
almost everywhere equal to a function in $C([0,T])$. Then:
\begin{itemize}
\item[(i)] $u$  is almost everywhere equal to a function  $\Tilde{u}\in C([0,T])$.

\item[(ii)] ${}_tD_{T}^\alpha u(t)$ exists for any $t\in [0,T]$ 
and ${}_tD_{T}^\alpha {u}\in C([0,T])$.
\end{itemize}
\end{lemma}

\begin{proof}[Proof of Theorem \ref{rt1}]
 Suppose that $u$ is the weak
solution of Problem \eqref{ip0}. Set $g(t)=\lambda
f(t,u(t))-u(t)$. By Definition of weak solution, one has
$$
\int_0^T{}_0 D_t^{\alpha} u(t)_0 D_t^{\alpha}v(t)dt
=\int_0^Tg(t)v(t)dt\quad \forall v\in E_0^\alpha.
$$
Thus from definition $\ref{rd1}$, we have 
$g(t)={}_t \overline{D}_T^{\alpha}{}_0 D_t^{\alpha} u(t)$ and from Remark
\ref{rr1}, we can conclude 
$g(t)={}_t \overline{D}_T^{\alpha}{}_0 \overline{D}_t^{\alpha} u(t)$.

From Theorem \ref{pt2}, $g\in C([0,T])$. Then Lemma \ref{rl5} implies that
${}_t D_T^{\alpha}{}_0 \overline{D}_t^{\alpha} u(t)$ exists for any 
$t\in [0,T]$, ${}_t D_T^{\alpha}{}_0 \overline{D}_t^{\alpha} u(t)\in C([0,T])$ 
and ${}_0 \overline{D}_t^{\alpha}u$ is almost everywhere equal to an 
element of C([0,T]). Then from Lemma $\ref{rl3}(ii)$, we have 
${}_0 D_t^{\alpha}u$ exists for any $t\in [0,T]$ and Remark 
\ref{rr1} yields ${}_0 \overline{D}_t^{\alpha}u={}_0 D_t^{\alpha}u$ 
a.e. on $[0,T]$. Hence, we  conclude that ${}_t D_T^{\alpha}{}_0 D_t^{\alpha} u(t)$ 
exists for any $t\in [0,T]$.

 Since $g$ and ${}_t D_T^{\alpha}{}_0 D_t^{\alpha} u$ are almost everywhere 
equal and they are continuous, we have
 $$
{}_t D_T^{\alpha}{}_0 D_t^{\alpha} u(t)=g(t)\quad \forall  t\in [0,T],
$$
which completes the proof.
\end{proof}

\subsection*{Acknowledgements}
The authors are grateful to the anonymous referees for their valuable
comments and suggestions.

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\end{document}