\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 184, pp. 1--13.\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-2014/184\hfil Regularity of mild solutions] {Regularity of mild solutions to fractional Cauchy problems with Riemann-Liouville fractional derivative} \author[Y.-N. Li, H.-R. Sun \hfil EJDE-2014/184\hfilneg] {Ya-Ning Li, Hong-Rui Sun} % in alphabetical order \address{Ya-Ning Li \newline School of Mathematics and Statistics, Lanzhou University\\ Lanzhou, Gansu 730000, China.\newline College of Mathematics \& Statistics, Nanjing University of Information Science \& Technology, Nanjing, 210044, China} \email{liyn08@lzu.edu.cn} \address{Hong-Rui Sun \newline School of Mathematics and Statistics, Lanzhou University\\ Lanzhou, Gansu 730000, China} \email{hrsun@lzu.edu.cn} \thanks{Submitted November 29, 2013. Published August 29, 2014.} \subjclass[2000]{34G10} \keywords{Fractional drivative; Cauchy problem; Mittag-Leffler function; \hfill\break\indent mild solution} \begin{abstract} As an extension of the fact that a sectorial operator can determine an analytic semigroup, we first show that a sectorial operator can determine a real analytic $\alpha$-order fractional resolvent which is defined in terms of Mittag-Leffler function and the curve integral. Then we give some properties of real analytic $\alpha$-order fractional resolvent. Finally, based on these properties, we discuss the regularity of mild solution of a class of fractional abstract Cauchy problems with Riemann-Liouville fractional derivative. \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 are widely and efficiently used to describe many phenomena arising in viscoelasticity, fractal, porous media, economic and science. More details on this theory and its applications can be found in \cite{AL, DH, G, H, K, Ma, Me, M, Po, T, ZhouYBook}. Recently, fractional abstract Cauchy problems have attracted much attention due to their wide application. Bajlekova \cite{B} defined a solution operator which extends the classical semigroup to study the fractional abstract Cauchy problem. Under the condition that the coefficient operator is the generator of a solution operator, some authors got the existence and uniqueness of mild solution of the inhomogeneous $\alpha$-order abstract Cauchy problem \cite{He,LC,LP2,LP}. Under the condition that the coefficient operator generates a $C_0$-semigroup, there is another tool to deal with the fractional abstract Cauchy problem, it is a new operator described by the $C_0$-semigroup and the probability density function. For more details, we refer to \cite{E,E0,E1,WZ,ZJ,ZJ1,ZYTMNA13}. However, these papers considered the fractional abstract Cauchy problem only in the Cupto's sense. Heymans and Podlubny \cite{Hey} showed that in some examples from the field of viscoelasticity, it is possible to attribute physical meaning to initial conditions expressed in terms of Riemann-Liouville fractional derivative or integral. Li, Peng and Jia \cite {LP1} developed an operator theory to study fractional abstract Cauchy problem with Riemann-Liouville fractional derivative. They proved that a homogeneous $\alpha$-order Cauchy problem is well posed if and only if its coefficient operator is the generator of an $\alpha$-order fractional resolvent, and gave sufficient conditions to guarantee the existence and uniqueness of weak solutions and strong solutions of an inhomogeneous $\alpha$-order Cauchy problem. On the other hand, it is well known that a sectorial operator can determine an analytic semigroup. Thus, it is natural to ask whether a sectorial operator can determine a real analytic $\alpha$-order fractional resolvent. Our first aim in this paper is to show that a sectorial operator of angle $\theta\in [0,(1-\frac{\alpha}{2})\pi)$ determines a real analytic $\alpha$-order fractional resolvent $\{T_{\alpha}(t)\}_{t\geq0}$ which is defined in terms of Mittag-Leffler function and the curve integral. We also present some properties of $\{T_{\alpha}(t)\}_{t\geq0}$. Our second purpose is to study the regularity of mild solution of an inhomogeneous $\alpha$-order abstract Cauchy problem. To the best of the authors' knowledge, the regularity of mild solution of fractional abstract Cauchy problem is a subject that has not been treated in the literature. So, in this paper, we will fill the gap in this area. We discuss the regularity of mild solution of the problem \begin{equation} \begin{gathered} D_t^{\alpha}u(t)+Au(t)=f(t),\quad t\in(0,T],\\ (g_{2-\alpha}\ast u)(0)=0,\quad (g_{2-\alpha}\ast u)'(0)=x, \end{gathered}\label{50} \end{equation} where $1<\alpha<2$, $A$ is a sectorial operator of angle $\theta\in [0,(1-\frac{\alpha}{2})\pi)$, $D_t^{\alpha}$ is the $\alpha$-order Riemann-Liouville fractional derivative operator, $g_{2-\alpha}(t)=\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}$ for $t>0$ and $g_{2-\alpha}(t)=0$ for $t\leq0$, $f:[0,T]\to X$, $X$ is a Banach space, $x \in X$. We prove that if $f\in L^p((0,T);X)$ with $p\in(\frac{1}{\alpha},\frac{1}{\alpha-1})$ then the mild solution of \eqref{50} is H\"{o}lder continuous on $(\varepsilon,T]$ for every $\varepsilon>0$. We also show that, the H\"older continuity of $f$ ensures that the mild solution $u$ of \eqref{50} is a classical solution and $Au,\ D_t^{\alpha}u$ is H\"older continuous. The rest of this paper is organized as follows. In Section 2, we provide some preliminaries of the fractional calculus and the Mittag-Leffler function. In Section 3, we introduce an operator family $\{T_{\alpha}(t)\}_{t\geq0}$ and analyze its properties. The regularity of mild solution of \eqref{50} is established in Section 4. \section{Preliminaries} Throughout this paper, let $X$ be a Banach space, $B(X)$ denotes the space of all bounded linear operators from $X$ to $X$. If $A$ is a closed linear operator, $\rho(A)$ and $\sigma(A)$ denote the resolvent set and the spectral set of $A$ respectively, $R(\lambda, A)=(\lambda I-A)^{-1}$ denotes the resolvent operator of $A$. $L^1(\mathbb{R}^+, X)$ denotes the Banach space of $X$-valued Bochner integrable functions. For convenience, we recall the following known definitions. By $\ast$ we denote the convolution of functions $(f\ast g)(t)=\int_0^tf(t-\tau)g(\tau)d\tau,\ t\geq0$. Let $g_{\alpha}(\alpha>0)$ denotes the function $$ g_{\alpha}(t)=\begin{cases} \frac{t^{\alpha-1}}{\Gamma(\alpha)},& t>0,\\ 0,& t\leq 0, \end{cases} $$ and $g_0(t)=\delta_0(t)$, the Dirac delta function. The Riemann-Liouville fractional integral of order $\alpha>0$ of $f$ is defined by $$ J_t^{\alpha}f(t)=(g_{\alpha}\ast f)(t). $$ The Riemann-Liouville fractional derivative of order $\alpha>0$ of $f$ can be written as $$ D_t^{\alpha}f(t) =\frac{d^m}{dt^m}J_t^{m-\alpha}f(t), $$ where $m$ is the smallest integer greater than or equal to $\alpha$. For more details about fractional calculus, we refer to \cite{K,M,Po,ZhouYBook}. The Mittag-Leffler function is defined by $$ E_{\alpha, \beta}(z)=\sum_{n=0}^{\infty}\frac{z^n} {\Gamma(\alpha n+\beta)},\quad z,\beta\in\mathbb{C},\;\operatorname{Re} \alpha>0. $$ The Mittag-Leffler function has the following properties (see \cite{K}): \begin{gather} \int_0^{\infty}e^{-\lambda t}t^{\beta-1}E_{\alpha, \beta}(\mu t^{\alpha})dt=\frac{\lambda^{\alpha-\beta}} {\lambda^{\alpha}-\mu}, \quad \operatorname{Re}\lambda>|\mu|^{1/\alpha}, \label{19} \\ \frac{d^n}{dt^n}(t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha}))= t^{\alpha-n-1}E_{\alpha,\alpha-n}(\mu t^{\alpha}), \quad n\in\mathbb{Z}^+.\label{4} \end{gather} The following lemma gives asymptotic formulae for the Mittag-Leffler functions. \begin{lemma}[{\cite[Theorem 1.4]{Po}}] \label{lem2} If $0<\alpha<2$, $\beta$ is an arbitrary real number, then for an arbitrary integer $N>1$, \begin{equation} E_{\alpha,\beta}(z) =-\sum_{n=1}^{N-1}\frac{z^{-n}}{\Gamma(\beta-\alpha n)}+O(|z|^{-N}),\quad \frac{\pi\alpha}{2}<|\arg z|\leq\pi, \label{9} \end{equation} as $|z|\to\infty$. \end{lemma} \begin{remark} \label{rem1} \rm Since $\frac{1}{\Gamma(-n)}=0$, $n=0,1,2,\ldots$, from \eqref{9}, we know if $\beta-\alpha=-n$, $(n=0,1,2,\ldots)$, \begin{equation} |E_{\alpha,\beta}(z)|\leq \frac{C}{1+|z|^2},\quad \frac{\pi\alpha}{2}<|\arg z|\leq\pi,\label{37} \end{equation} where $C$ is a real constant. \end{remark} Now, we present introduction to sectorial operators. \begin{definition}[{\cite[Definition 1.2.1]{C}}] \rm Let $A$ be a densely defined closed linear operator on Banach space $X$, then $A$ is called a sectorial operator of angle $\omega\in[0,\pi)$ ($A\in \operatorname{Sect}(w)$, in short) if \begin{itemize} \item[(1)] $\sigma(A)\subseteq\overline{\Sigma_{\omega}}$, where \[ \Sigma_{\omega}:=\begin{cases} \{z\in\mathbb{C}:z\neq0\text{ and } |\arg z|<\omega\},& \omega>0,\\ (0,\infty), & \omega=0, \end{cases} \] \item[(2)] for every $\omega'\in(\omega,\pi)$, $\sup\{\|zR(z,A)\|:z\in\mathbb{C}\setminus \overline{\Sigma_{\omega'}}\}<\infty$. \end{itemize} \end{definition} For a closed linear operator $A$ on a Banach space $X$, recall the following statement. \begin{lemma}[{\cite[Proposition 1.1.7]{AB}}] \label{lem3} Let $A$ be a closed linear operator on $X$ and $I$ be an interval in $\mathbb{R}$. Let $f:I\to X$ be Bochner integrable. Suppose that $f(t)\in D(A)$ for $t\in I$ and $Af:I\to X$ is Bochner integrable. Then $\int_If(t)dt\in D(A)$ and $A\int_If(t)dt=\int_IAf(t)dt$. \end{lemma} The following definition is a direct consequence of \cite[Definition 3.1 and Theorem 3.12]{LP1}. \begin{definition} \rm Let $A$ be a closed linear operator defined on $X$ and $1\leq \alpha \leq 2$. A family $\{T_{\alpha}(t)\}_{t\geq0}\subset B(X)$ is called an $\alpha$-order fractional resolvent generated by $A$, if for every $t\geq0$, $T_{\alpha}(t)$ is strongly continuous and there exists $\omega\in \mathbb{R}$ such that $\{\lambda^{\alpha}:\operatorname{Re}\lambda>\omega\}\subset\rho(A)$ and $(\lambda^{\alpha}-A)^{-1}x=\int_0^{\infty}e^{-\lambda t}T_{\alpha}(t)x\,dt$, $\operatorname{Re}\lambda>\omega$, $x\in X$. \end{definition} $\{T_{\alpha}(t)\}_{t\geq0}$ has the following property \cite[Proposition 3.7]{LP1}: If $-A$ is the generator of $\{T_{\alpha}(t)\}_{t\geq0}$, then for every $t\geq0$ and $x\in X$, $(g_{\alpha}\ast T_{\alpha})(t)x\in D(A)$, and \begin{equation} T_{\alpha}(t)x=g_{\alpha}(t)x-A(g_{\alpha}\ast T_{\alpha})(t)x. \label{21} \end{equation} Below the letter $C$ denotes various positive constants, and $C_{\alpha}$ denote various positive constants depending on $\alpha$. \section{The operator $T_{\alpha}(t)$} For the rest of this article, let $1<\alpha<2$, $A\in \operatorname{Sect}(\theta)$ with $\theta\in [0,(1-\frac{\alpha}{2})\pi)$ and $0\in\rho(A)$. Inspired by the expression of an analytic semigroup determined by a sectorial operator $A$, we introduce an operator family $\{T_{\alpha}(t)\}_{t\geq0}$ by \begin{equation} T_{\alpha}(t)= \frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha})(\mu I+A)^{-1}d\mu, \label{1} \end{equation} where the integral path $\Gamma_{\pi-\theta}:=\{\mathbb{R}^+e^{i(\pi-\theta)}\} \cup\{\mathbb{R}^+e^{-i(\pi-\theta)}\}$ is oriented counter clockwise. First, we show some basic properties of $\{T_{\alpha}(t)\}_{t\geq0}$. \begin{theorem}\label{th1} For every $t\geq0$, $T_{\alpha}(t)$ is well defined and $\{T_{\alpha}(t)\}_{t\geq0}$ is a real analytic $\alpha$-order fractional resolvent. Moreover, there exists a constant $C_{\alpha}$ such that \begin{equation} \|T_{\alpha}(t)\|\leq C_{\alpha}t^{\alpha-1},\quad t\geq0. \label{18} \end{equation} \end{theorem} \begin{proof} $A\in \operatorname{Sect}(\theta)$ implies that $\Sigma_{\pi-\theta}\subset \rho (-A)$ and \begin{equation} \|(\mu I+A)^{-1}\|\leq \frac{C}{|\mu|},\quad \mu\in\Gamma_{\pi-\theta}\setminus \{0\},\label{3} \end{equation} which combines with Remark \ref{rem1}, we can get that, for every $t\geq0$, $T_{\alpha}(t)$ is well defined. For $\mu\in\Gamma_{\pi-\theta}$, since $(\mu I+A)^{-1}$ is a bounded linear operator, it is easy to see that $T_{\alpha}(t)$ is also a bounded linear operator. Now, we show that $\{T_{\alpha}(t)\}_{t\geq0}$ is an $\alpha$-order fractional resolvent generated by $-A$. We first show that, for every $t\geq0$, $T_{\alpha}(t)$ is a strongly continuous operator. Fix $t_0\geq0$, then for $t>0$, $x\in X$, we have $$ T_{\alpha}(t)x-T_{\alpha}(t_0)x=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}(t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha})-t_0^{\alpha-1}E_{\alpha,\alpha}(\mu t_0^{\alpha}))(\mu I+A)^{-1}xd\mu. $$ Then by the continuity of $t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha})$ and the dominated convergence theorem, we know that $\lim_{t\to t_0}T_{\alpha}(t)x=T_{\alpha}(t_0)x$. Let $\theta_0\in (\frac{\pi}{2}, \frac{\pi-\theta}{\alpha})$, $\varrho>0$, and \begin{equation} l_{\theta_0}:=\{r e^{-i\theta_0},\varrho\leq r<\infty\}\cup\{\varrho e^{i\varphi}, |\varphi|<\theta_0\}\cup\{re^{i\theta_0},\varrho\leq r<\infty\} \label{24} \end{equation} be oriented counter clockwise. Then for $\lambda\in l_{\theta_0}$, $\lambda^{\alpha}\in\Sigma_{\pi-\theta}\subset\rho(-A)$, hence $\{\lambda^{\alpha}:Re \lambda>\varrho\}\subset \rho(-A)$. In view of \eqref{19}, we know that \begin{equation} t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha})=\frac{1}{2\pi i} \int_{l_{\theta_0}}e^{\lambda t}(\lambda^{\alpha}-\mu)^{-1}d\lambda,\ \mu\in\Gamma_{\pi-\theta}. \label{32} \end{equation} For $x\in X$, from Fubini's theorem, \eqref{32} and the Cauchy's integral formula, we see that \begin{equation} \begin{aligned} T_{\alpha}(t)x &=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}} t^{\alpha-1}E_{\alpha,\alpha}(\mu t^{\alpha})(\mu I+A)^{-1}xd\mu\\ &=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}\frac{1}{2\pi i} \int_{l_{\theta_0}}e^{\lambda t}(\lambda^{\alpha}-\mu)^{-1}d\lambda(\mu I+A)^{-1}xd\mu\\ &=\frac{1}{2\pi i}\int_{l_{\theta_0}}e^{\lambda t}\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}} (\lambda^{\alpha}-\mu)^{-1}(\mu I+A)^{-1}xd\mu d\lambda\\ &=\frac{1}{2\pi i}\int_{l_{\theta_0}}e^{\lambda t}(\lambda^{\alpha}I+A)^{-1}xd\lambda. \end{aligned} \label{39} \end{equation} Then taking Laplace transform on both sides, we obtain \begin{equation} \label{add} (\lambda^{\alpha}I+A)^{-1}x=\int_0^{\infty}e^{-\lambda t}T_{\alpha}(t)x\,dt,\quad \operatorname{Re}\lambda>\varrho,\; x\in X. \end{equation} Next, we prove that the estimate \eqref{18} holds. It is clear that $T_{\alpha}(0)=0$. For $t>0$, in view of \eqref{39} and \eqref{3}, we deduce \begin{align*} \|T_{\alpha}(t)\| &=\|\frac{1}{2\pi i}\int_{l_{\theta_0}}e^{\lambda t}(\lambda^{\alpha}I+A)^{-1}d \lambda\|\\ &=\frac{1}{2\pi} \|\int_{l'_{\theta_0}}e^{\mu} ((\frac{\mu}{t})^{\alpha}I+A)^{-1}\frac{1}{t}d\mu\|\\ &\leq \frac{C}{2\pi}\int_{l'_{\theta_0}}|e^{\mu}| \frac{t^{\alpha-1}}{|\mu|^{\alpha}}|d\mu|=C_{\alpha}t^{\alpha-1}. \end{align*} Finally, we verify that $T_{\alpha}(t)$ is real analytic. From the dominated convergence theorem and \eqref{4}, we have, for $n\in \mathbb{N}^+$, \begin{align*} T_{\alpha}^{(n)}(t) &=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}t^{\alpha-n-1}E_{\alpha,\alpha-n}(\mu t^{\alpha})(\mu I+A)^{-1}d\mu\\ &=\frac{1}{2\pi i}\int_{\Gamma'_{\pi-\theta}}t^{\alpha-n-1} E_{\alpha,\alpha-n}(\xi)(\frac{\xi}{t^{\alpha}} I+A)^{-1}\frac{1}{t^{\alpha}}d\xi. \end{align*} This combined with \eqref{3}, yields \begin{equation} \|T_{\alpha}^{(n)}(t)\|\leq C_{\alpha}t^{\alpha-n-1},\quad t\geq0.\label{8} \end{equation} Let $\tilde{c}:=\inf_{n\in\mathbb{N}^+}\{C_{\alpha}^{-\frac{1}{n}}\}$, where $C_{\alpha}$ is given in \eqref{8}. For fixed $z\in\mathbb{R}^+$, denote $\tilde{z}:=\inf_{n\in\mathbb{N}^+}\{z^{1+\frac{1-\alpha}{n}}\}$. Choose $|t-z|\leq K\tilde{c}\tilde{z}$, $00$ and $x\in X$, we have $T_{\alpha}(t)x\in D(A)$ and $\|AT_{\alpha}(t)\|\leq\frac{C}{t}$. \end{theorem} \begin{proof} From $A(\lambda^{\alpha}I+A)^{-1}=I-\lambda^{\alpha}(\lambda^{\alpha}I+A)^{-1}$, for $t> 0$ and $x\in X$, we have \begin{align*} &\int_{l_{\theta_0}}e^{\lambda t}A(\lambda^{\alpha}I+A)^{-1}x\,d\lambda\\ &=\int_{l_{\theta_0}}e^{\lambda t}xd\,\lambda -\int_{l_{\theta_0}}e^{\lambda t}\lambda^{\alpha}(\lambda^{\alpha}I+A)^{-1}x\,d\lambda\\ &=\int_{l'_{\theta_0}}e^{\mu}\frac{1}{t}xd\mu -\int_{l'_{\theta_0}}e^{\mu}(\frac{\mu}{t})^{\alpha} ((\frac{\mu}{t})^{\alpha}I+A)^{-1}\frac{1}{t}x\,d\mu, \end{align*} where $l_{\theta_0}$ is given by $\eqref{24}$. Since $\theta_0<\frac{\pi-\theta}{\alpha}$, for $\mu\in l'_{\theta_0}$, we have $\mu^{\alpha}\in \Sigma_{\pi-\theta}\subset\rho(-A)$, and \begin{equation} \|((\mu/t)^{\alpha}I+A)^{-1}\| \leq\frac{Ct^{\alpha}}{|\mu|^{\alpha}}. \end{equation} Consequently, \begin{equation} \|\int_{l_{\theta_0}}e^{\lambda t}A(\lambda^{\alpha}I+A)^{-1}xd\lambda\| \leq\frac{C}{t} \int_{l'_{\theta_0}}|e^{\mu}||d\mu|\leq\frac{C}{t}. \label{20} \end{equation} Thus, by \eqref{39}, \eqref{20}, the closeness of $A$ and Lemma \ref{lem3}, we conclude that for every $x\in X$ and $t>0$, $T_{\alpha}(t)x\in D(A)$ and $\|AT_{\alpha}(t)\|\leq\frac{C}{t}$. \end{proof} \section{Main results} In this section, we apply the theory developed in Section 3 to discuss the regularity of mild solution of the following linear inhomogeneous fractional Cauchy problem \begin{equation} \begin{gathered} D_t^{\alpha}u(t)+Au(t)=f(t),\quad t\in(0,T],\\ (g_{2-\alpha}\ast u)(0)=0,\quad (g_{2-\alpha}\ast u)'(0)=x, \end{gathered} \label{5} \end{equation} where $f\in L^1((0,T);X)$ and $x \in X$. To present definition of mild solution of problem \eqref{5}, we give the following lemmas. \begin{lemma}\label{lema1} Suppose $u\in C([0,T]; X)$ such that $(g_{2-\alpha} \ast u)\in C^2((0,T]; X), u(t)\in D(A)$ for $t\in[0, T], Au\in L^1((0,T);X)$ and $u$ satisfies \eqref{5}. Then \begin{equation} u(t)=T_{\alpha}(t)x+ \int_0^tT_{\alpha}(t-s) f(s)ds. \label{7} \end{equation} \end{lemma} \begin{proof} If $u$ satisfies the assumptions, we can write $u$ as \begin{equation} u(t)=g_{\alpha}(t)x-A(g_{\alpha}\ast u)(t)+(g_{\alpha}\ast f)(t), \quad t\in[0,T]. \label{6} \end{equation} Applying the Laplace transform to \eqref{6}, then, for $\lambda>0$, $$ \hat{u}(\lambda)=\lambda^{-\alpha}x -\lambda^{-\alpha}A\hat{u}(\lambda) +\lambda^{-\alpha}\hat{f}(\lambda); $$ that is \begin{equation} \hat{u}(\lambda)=(\lambda^{\alpha}I+A)^{-1}x +(\lambda^{\alpha}I+A)^{-1}\hat{f}(\lambda),\quad \lambda>0.\label{45} \end{equation} Then taking inverse Laplace transform to \eqref{45} and by \eqref{add}, we obtain the conclusion. \end{proof} \begin{lemma}\label{lem4.2} If $f\in L^1((0,T);X)$, then the integral $\int_0^tT_{\alpha}(t-s) f(s)\,ds$ exists and defines a continuous function. \end{lemma} \begin{proof} Since $f\in L^1((0,T);X)$, $T_{\alpha}(t)\in B(X)$ for $t\in(0,T)$, by \cite[Theorem 1.3.4]{P}, we know that $(T_{\alpha}\ast f)(t)=\int_0^tT_{\alpha}(t-s) f(s)\,ds$ exists and defines a continuous function. \end{proof} \begin{definition}\label{de1} \rm The function $u\in C([0,T],X)$ given by $$ u(t)=T_{\alpha}(t)x+\int_0^tT_{\alpha}(t-s) f(s)ds $$ is called a mild solution of the Cauchy problem \eqref{5}. \end{definition} By Definition \ref{de1} and Lemma \ref{lema1}, for $f\in L^1((0,T);X)$, we know the Cauchy problem \eqref{5} has a unique mild solution. \begin{definition} \rm A function $u\in C([0,T],X)$ is called a classical solution of \eqref{5} if $D_t^{\alpha}u\in C((0,T],X)$, and for all $t\in(0,T]$, $u(t)\in D(A)$ and satisfies \eqref{5}. \end{definition} \begin{theorem} Let $u$ be the mild solution of \eqref{5}. If $f\in L^p((0,T);X)$ with $\frac{1}{\alpha}0$. \end{theorem} \begin{proof} By \eqref{8}, we have $\|T'_{\alpha}(t)\|\leq C_{\alpha}t^{\alpha-2}$, then from the mean value theorem, we know that $T_{\alpha}(t)x$ is Lipschitz continuous on $[\varepsilon,T]$ for every $\varepsilon>0$. If $\frac{1}{\alpha}0$ and $t\in[0,T-h]$, we have \begin{align*} v(t+h)-v(t) &=\int_0^{t+h}T_{\alpha}(t+h-s)f(s)ds-\int_0^{t}T_{\alpha}(t-s)f(s)ds\\ &=\int_t^{t+h}T_{\alpha}(t+h-s)f(s)ds +\int_0^{t}(T_{\alpha}(t+h-s)-T_{\alpha}(t-s))f(s)ds\\ &=I_1+I_2. \end{align*} By \eqref{18} and $p>1/\alpha$, we have \begin{align*} \|I_1\| &\leq C_{\alpha}\int_t^{t+h}(t+h-s)^{\alpha-1}\|f(s)\|ds\\ &\leq C_{\alpha}\Big(\int_t^{t+h}(t+h-s)^{\frac{p(\alpha-1)} {p-1}}ds\Big)^{\frac{p-1}{p}}\|f\|_{L^p}\\ &\leq C_{\alpha}\|f\|_{L^p}h^{\frac{\alpha p-1}{p}}. \end{align*} To estimate $I_2$, we use that \eqref{18} implies $$ \|T_{\alpha}(t+h)-T_{\alpha}(t)\|\leq C_{\alpha}T^{\alpha-1}. $$ On the other hand, from the mean value theorem and \eqref{8}, we obtain $$ \|T_{\alpha}(t+h)-T_{\alpha}(t)\|\leq C_{\alpha}t^{\alpha-2}h. $$ Therefore, \begin{equation} \|T_{\alpha}(t+h)-T_{\alpha}(t)\|\leq\mu(h,t):= C_{\alpha}\min\{T^{\alpha-1},\ t^{\alpha-2}h\}.\label{11} \end{equation} Using \eqref{11} and the H\"older's inequality, we have \begin{align*} \|I_2\| &\leq C_{\alpha}\int_0^t\mu(h,t-s)\|f(s)\|ds\\ &\leq C_{\alpha}\|f\|_{L^p}\Big(\int_0^t\mu(h,t-s) ^{\frac{p}{p-1}}ds\Big)^{\frac{p-1}{p}}\\ &= C_{\alpha}\|f\|_{L^p}\Big(\int_0^t\mu(h,\tau) ^{\frac{p}{p-1}}d\tau\Big)^{\frac{p-1}{p}}\\ &\leq C_{\alpha}\|f\|_{L^p}\Big(\int_0^{\infty} \mu(h,\tau)^{\frac{p}{p-1}}d\tau\Big)^{\frac{p-1}{p}}\\ &=C_{\alpha}\|f\|_{L^p}T^{\alpha-1}h +C_{\alpha}\|f\|_{L^p}\Big(\int_h^{\infty}\tau ^{\frac{p(\alpha-2)}{p-1}}ds\Big)^{\frac{p-1}{p}}h\\ &=C_{\alpha}\|f\|_{L^p}T^{\alpha-1}h +C_{\alpha}\|f\|_{L^p}h^{\frac{p\alpha-1}{p}}\\ &\leq C_{\alpha}\|f\|_{L^p}h^{\frac{\alpha p-1}{p}}. \end{align*} \end{proof} \begin{theorem}\label{th2} Suppose $f\in C^{\gamma}([0,T];X)$ for $\gamma\in(0,1)$; that is, there is a constant $k>0$ such that $$ \|f(t)-f(s)\|\leq k|t-s|^{\gamma},\ 00$ and $t\in(0,T-h]$, we have \begin{equation} \begin{aligned} AI_1(t+h)-AI_1(t) &=\int_0^tA[T_{\alpha}(t+h-s)-T_{\alpha}(t-s)](f(s)-f(t))ds\\ &\quad +\int_0^{t}AT_{\alpha}(t+h-s)(f(t)-f(t+h))ds\\ &\quad +\int_t^{t+h}AT_{\alpha}(t+h-s)(f(s)-f(t+h))ds\\ &=h_1+h_2+h_3. \end{aligned} \label{25} \end{equation} For $h_1$, $$ \lim_{h\to0}AT_{\alpha}(t+h-s)(f(s)-f(t))=AT_{\alpha}(t-s)(f(s)-f(t)), $$ and from Theorem \ref{th7}, we know that $$ \|AT_{\alpha}(t+h-s)(f(s)-f(t))\| \leq C(t+h-s)^{-1}(t-s)^{\gamma}\leq C(t-s)^{\gamma-1}\in L^1(0,t). $$ Thus, by means of the dominated convergence theorem, we obtain that $h_1\to0$ as $h\to0$. For $h_2$, we have \begin{align*} \|h_2\| &=\|\int_0^{t}AT_{\alpha}(t+h-s)(f(t)-f(t+h))\,ds\|\\ &\leq C\int_0^t(t+h-s)^{-1}h^{\gamma}\,ds\\ &=C(\ln(t+h)-\ln h)h^{\gamma}, \end{align*} so, $h_2\to0$ as $h\to0$. Also $$ \|h_3\|\leq C\int_t^{t+h}(t+h-s)^{-1}(t+h-s)^{\gamma}ds =\frac{Ch^{\gamma}}{\gamma}\to0 \quad\text{as } h\to0. $$ Consequently, $Av\in C((0,T];X)$. It is easy to see that $ (g_{2-\alpha}\ast v)(0)=0$, $(g_{2-\alpha}\ast v)'(0)=0$. \end{proof} \begin{lemma}\label{lem1} Suppose $f\in C^{\gamma}([0,T];X)$ for $\gamma\in(0,1)$, denote $$ I_1(t):=\int_0^tT_{\alpha}(t-s)(f(s)-f(t))ds,\quad t\in(0,T], $$ then $I_1(t)\in D(A)$ for $0\leq t\leq T$ and $AI_1\in C^{\gamma}([0,T];X)$. \end{lemma} \begin{proof} The fact that $I_1(t)\in D(A)$ for $0\leq t\leq T$ is an immediate consequence of the proof of Theorem \ref{th2}, so we only need to prove the H\"older continuity of $AI_1(t)$. From the dominated convergence theorem and \eqref{4}, we have \begin{align*} &\frac{d}{dt}AT_{\alpha}(t)\\ &=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}t^{\alpha-2} E_{\alpha,\alpha-1}(\mu t^{\alpha})A(\mu I+A)^{-1}d\mu\\ &=\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}t^{\alpha-2} E_{\alpha,\alpha-1}(\mu t^{\alpha})d\mu -\frac{1}{2\pi i}\int_{\Gamma_{\pi-\theta}}t^{\alpha-2} E_{\alpha,\alpha-1}(\mu t^{\alpha})\mu(\mu I+A)^{-1}d\mu\\ &=\frac{1}{2\pi i}\int_{\Gamma'_{\pi-\theta}}t^{\alpha-2} E_{\alpha,\alpha-1}(\xi) \frac{1}{t^{\alpha}}d\xi -\frac{1}{2\pi i}\int_{\Gamma'_{\pi-\theta}}t^{\alpha-2} E_{\alpha,\alpha-1}(\xi) \frac{\xi}{t^{\alpha}}(\frac{\xi}{t^{\alpha}}I+A)^{-1}\frac{1}{t^{\alpha}}d\xi. \end{align*} In view of \eqref{3}, we deduce that \begin{equation} \|\frac{d}{dt}AT_{\alpha}(t)\|\leq C_{\alpha}t^{-2}, \quad 00$ and $t\in[0,T-h]$, from \eqref{25}, we know that \begin{align} AI_1(t+h)-AI_1(t)=h_1+h_2+h_3. \label{15} \end{align} From $f\in C^{\gamma}([0,T];X)$ and \eqref{12}, it follows that \begin{equation} \begin{aligned} \|h_1\| &\leq \int_0^{t}\|AT_{\alpha}(t+h-s)-AT_{\alpha}(t-s)\|\|f(s)-f(t)\|\,ds\\ &\leq C_{\alpha}h\int_0^t(t+h-s)^{-1}(t-s)^{\gamma-1}\,ds\\ &=C_{\alpha}h\int_0^t(s+h)^{-1}s^{\gamma-1}\,ds\\ &\leq C_{\alpha}\int_0^h\frac{h}{s+h}s^{\gamma-1}ds+C_{\alpha}\int_h^{\infty} \frac{s^{\gamma-1}}{s+h}h\,ds\\ &\leq C_{\alpha}\int_0^hs^{\gamma-1}\,ds+ C_{\alpha}\int_h^{\infty}s^{\gamma-2}h\,ds =C_{\alpha}h^{\gamma}. \label{16} \end{aligned} \end{equation} For $h_2$, by Theorem \ref{th7} and the mean value theorem, we have \begin{equation} \begin{aligned} \|h_2\| &\leq \int_0^t\|AT_{\alpha}(t+h-s)\|\|f(t)-f(t+h)\|\,ds\\ &\leq C\int_0^t(t+h-s)^{-1}\,ds\, h^{\gamma} = C\int_h^{t+h}s^{-1}\,ds\, h^{\gamma}\\ &=C\frac{t}{\theta t+h}h^{\gamma}\leq\frac{C}{\theta}h^{\gamma}, \end{aligned} \label{13} \end{equation} where $\theta\in(0,1)$. For $h_3$, it follows from Theorem \ref{th7} and the assumption on $f$, we see that \begin{equation} \begin{aligned} \|h_3\| &\leq\int_t^{t+h}\|AT_{\alpha}(t+h-s)\|\|f(s)-f(t+h)\|\,ds\\ &\leq C\int_t^{t+h}(t+h-s)^{\gamma-1}\,ds\leq Ch^{\gamma}. \end{aligned} \label{14} \end{equation} Combining \eqref{15} with the estimates \eqref{16}, \eqref{13} and \eqref{14}, we obtain that $AI_1$ is H\"older continuous with exponent $\gamma$ on [0,T]. \end{proof} \begin{theorem} \label{thm4.6} Suppose $f\in C^{\gamma}([0,T];X)$ for $\gamma\in(0,1)$. If $u$ is a classical solution of the problem \eqref{5} on $[0,T]$, then \begin{itemize} \item[(i)] For every $\varepsilon>0$, $Au\in C^{\gamma}([\varepsilon,T];X)$ and $D_t^{\alpha}u(t)\in C^{\gamma}([\varepsilon,T];X)$. \item[(ii)] If $x\in D(A), f(0)=0$, then $Au$ and $D_t^{\alpha}u(t)$ are continuous on $[0,T]$. \item[(iii)] If $x=0, f(0)=0$, then $Au, D_t^{\alpha}u(t)\in C^{\gamma}([0,T];X)$. \end{itemize} \end{theorem} \begin{proof} (i) If $u$ is a classical solution of the initial value problem \eqref{5} on $[0,T]$, then $$ u(t)=T_{\alpha}(t)x+\int_0^tT_{\alpha}(t-s) f(s)ds=T_{\alpha}(t)x+v(t). $$ By \eqref{10}, we know that $AT_{\alpha}(t)x$ is Lipschitz continuous on $[\varepsilon,T]$ for every $\varepsilon>0$. So, it suffices to show that $Av \in C^{\gamma}([\varepsilon,T];X)$. As in Theorem \ref{th2}, we write $v(t)$ as $$ v(t)=I_1(t)+I_2(t) =\int_0^t T_{\alpha}(t-s)(f(s)-f(t))ds+\int_0^tT_{\alpha}(t-s)f(t)ds, $$ for $00$. To this end, let $ h>0$ and $t\in[\varepsilon,T-h]$, then \begin{align*} AI_2(t+h)-AI_2(t) &=\int_0^{t+h}AT_{\alpha}(t+h-s)f(t+h)\,ds-\int_0^tAT_{\alpha}(t-s)f(t)\,ds\\ &=\int_0^{t+h}AT_{\alpha}(s)f(t+h)\,ds-\int_0^tAT_{\alpha}(s)f(t)\,ds\\ &=\int_0^{t+h}AT_{\alpha}(s)(f(t+h)-f(t))\,ds +\int_t^{t+h}AT_{\alpha}(s)f(t)\,ds. \end{align*} This combined with \eqref{17} yield \begin{align*} \| AI_2(t+h)-AI_2(t)\| &\leq C\|A(1\ast T_{\alpha})(t+h)\|h^{\gamma}+C\int_0^hs^{-1}ds\|f\|_{\infty}\\ &\leq Ch^{\gamma}+\frac{C}{\varepsilon}h\leq Ch^{\gamma}, \end{align*} where $\|f\|_{\infty}=\max_{0\leq t\leq T} \|f(t)\|$. (ii) If $x\in D(A)$, then $AT_{\alpha}(t)x\in C([0,T];X)$. By Lemma \ref{lem1} and $(i)$, we know that $AI_1 \in C^{\gamma}([0,T];X)$, $AI_2 \in C^{\gamma}([\varepsilon,T];X)$. We need to show that $AI_2$ is continuous at $t=0$. Since $f(0)=0$ and \eqref{17}, we have $\|AI_2(t)\|\leq \|(1\ast T_{\alpha})(t)\|\|f(t)\|\leq C \|f(t)\|\to 0$ as $t\to0$. This completes $(ii)$. (iii) We only to show that $AI_2\in C^{\gamma}([0,T];X)$. \begin{align*} &\|AI_2(t+h)-AI_2(t)\|\\ &\leq \|\int_0^{t+h}AT_{\alpha}(s)(f(t+h)-f(t))\,ds\| +\|\int_t^{t+h}AT_{\alpha}(s)f(t)\,ds\|\\ &\leq \|(1\ast AT_{\alpha})(t+h)\|\|f(t+h)-f(t)\|\,ds +\int_t^{t+h}\|AT_{\alpha}(s)\|\|f(t)-f(0)\|ds\\ &\leq Ch^{\gamma}+\int_t^{t+h}s^{-1}t^{\gamma}\,ds \leq Ch^{\gamma}+\int_t^{t+h}s^{\gamma-1}\,ds\\ &\leq Ch^{\gamma}+\int_0^{h}(t+s)^{\gamma-1}\,ds \leq Ch^{\gamma}+\int_0^{h}s^{\gamma-1}\,ds\\ &\leq Ch^{\gamma}. \end{align*} \end{proof} \subsection*{Acknowledgments} This research was supported by the program for New Century Excellent Talents in University (NECT-12-0246) and FRFCU (lzujbky-2013-k02). \begin{thebibliography}{99} \bibitem{AB} W. Arendt, C. Batty, M. Hieber, F. Neubrander; \emph{Vector-Valued Laplace Transforms and Cauchy Problems}, Monographs in Mathematics, 96. Birkh$\ddot{a}$user Verlag, Basel, 2001. \bibitem{AL} V. V. Anh, N. N. 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