\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 107, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/107\hfil Existence of mild solutions] {Existence of mild solutions for impulsive fractional-order semilinear evolution equations with nonlocal conditions} \author[A. Chauhan, J. Dabas \hfil EJDE-2011/107\hfilneg] {Archana Chauhan, Jaydev Dabas} % in alphabetical order \address{Archana Chauhan\newline Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad - 211 004, India} \email{archanasingh.chauhan@gmail.com} \address{Jaydev Dabas \newline Department of Paper Technology, IIT Roorkee, Saharanpur Campus, Saharanpur - 247001, India} \email{jay.dabas@gmail.com} \thanks{Submitted April 29, 2011. Published August 24, 2011.} \subjclass[2000]{34K05, 34A12, 34A37, 26A33} \keywords{Fractional order differential equation; nonlocal conditions; \hfill\break\indent contraction mapping; mild solution; impulsive conditions} \begin{abstract} In this work we consider a class of impulsive fractional-order semilinear evolution equations with a nonlocal initial condition. By means of solution operator and application of fixed point theorems we established the existence and uniqueness of a mild solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Recently fractional differential equations attracted many authors (see for instance \cite{da,mb,b1,dd,mm,mg,m,gmn,wj,yz} and references in these papers). Many phenomena in engineering, physics,continuum mechanics, signal processing, electromagnetics, economics and science describes efficiently by fractional order differential equations. Impulsive differential equations have become important in recent years as mathematical models of phenomena in both physical and social sciences (see for instance \cite{aa,b3,ZF,jl,mg,chen} and references in these papers). There has been a significant development in impulsive theory especially in the area of impulsive differential equations with fixed moments. In this article, we are concerned with the existence and uniqueness of the solution for the fractional order differential equation in a complex Banach space $X$, \begin{gather}\label{ME} \frac{d^{\alpha}}{dt^{\alpha}}x(t)+Ax(t) =f(t,x(t),x(a_1(t)),\dots,x(a_m(t))),\quad t\in J=[0,T],\; t\neq t_i,\\ x(0)+g(x)=x_0,\label{me}\\ \Delta x(t_i)=I_i(x(t_i^-)), \label{ME1} \end{gather} where $\frac{d^{\alpha}}{dt^\alpha}$ is Caputo's fractional derivative of order $0<\alpha<1$, $i=1,2,\dots,p$, $0=t_00,\;z\in \mathbb{C}, $$ where $H_a$ is a Hankel path, that is a contour which starts and ends at $-\infty$ and encircles the disc $|\mu|\leq|z|^{1\over \alpha}$ counter clockwise. It is an entire function which provides a generalization of several usual functions, for example: Exponent function: $E_{1,1}(z)=e^z$; cosine functions: $E_{2,1}(z^2)=\cosh(z)$ and $E_{2,1}(-z^2)=\cos(z);$ Sine functions: $zE_{2,2}(z^2)=\sinh(z)$ and $zE_{2,2}(-z^2)=\sin(z)$. The Laplace transform of the Mittag-Leffler function is given as: $$ L(t^{\beta-1}E_{\alpha,\beta}(-\rho^\alpha t^\alpha)) =\frac{\lambda^{\alpha-\beta}} {\lambda^\alpha+\rho^\alpha},\quad \operatorname{Re}\lambda>\rho^{1/\alpha},\;\rho>0. $$ To begin with the analysis we need some basic definitions and properties from the fractional calculus theory (see \cite{ip}). \begin{definition} \label{def2.1} \rm Caputo's derivative of order $\alpha$ for a function $f: [0,\infty)\to\mathbb{R}$ is defined as \[ {d^{\alpha}f(t)\over dt^{\alpha}}=\frac{1}{\Gamma(m-\alpha)} \int_0^t(t-s)^{n-\alpha-1}f^{(n)}(s)ds, \] for $n-1<\alpha0$ is given as $$ L\{D_t^\alpha f(t);\lambda\}=\lambda^\alpha\widehat{f}(\lambda) -\sum_{k=0}^{n-1}\lambda^{\alpha-k-1}f^{(k)}(0);\quad n-1<\alpha\leq n. $$ \end{definition} \begin{definition}[{\cite[Definition 2.3]{da}}]\label{def2.2} \rm Let $A$ be a closed and linear operator with domain $D(A)$ defined on a Banach space $X$ and $\alpha>0$. Let $\rho(A)$ be the resolvent set of $A$. We call $A$ the generator of an $\alpha-$resolvent family if there exists $\omega\geq 0$ and a strongly continuous function $S_\alpha:R_+\to L(X)$ such that $\{\lambda^\alpha:Re\lambda>\omega\}\subset\rho(A)$ and $$ (\lambda^\alpha I-A)^{-1}x=\int_0^\infty e^{-\lambda t}S_\alpha(t)x dt,\quad\operatorname{Re}\lambda>\omega,\; x\in X. $$ In this case, $S_\alpha(t)$ is called the $\alpha-$resolvent family generated by $A$. \end{definition} \begin{definition}[{\cite[Definition 2.1]{rp}}] \label{def2.3}\rm Let $A$ be a closed and linear operator with domain $D(A)$ defined on a Banach space $X$ and $\alpha>0$. Let $\rho(A)$ be the resolvent set of $A$, then we say that $A$ is the generator of a solution operator if there exists $\omega\geq 0$ and a strongly continuous function $S_\alpha:R_+\to L(X)$ such that $\{\lambda^\alpha:Re\lambda>\omega\}\subset\rho(A)$ and $$ \lambda^{\alpha-1}(\lambda^\alpha I-A)^{-1}x =\int_0^\infty e^{-\lambda t}S_\alpha(t)x dt,\quad\operatorname{Re}\lambda>\omega,\; x\in X. $$ In this case, $S_\alpha(t)$ is called the solution operator generated by $A$. \end{definition} The concept of solution operator is closely related to the concept of a resolvent family (see \cite[Chapter 1]{jp}). For more details on $\alpha$-resolvent family and solution operators, we refer to \cite{cl,jp} and the references therein. \section{Main results} In \cite{chen}, if $\alpha\in(0,1)$ and $A\in A^\alpha(\theta_0,\omega_0)$, then for any $x\in X$ and $t>0$, we have $$ \|T_\alpha(t)\|_{L(X)}\leq Me^{\omega t},\quad \|S_\alpha(t)\|_{L(X)}\leq Ce^{\omega t}(1+t^{\alpha-1}),\quad t>0,\; \omega>\omega_0. $$ Let $$ \widetilde{M}_T=\sup_{0\leq t\leq T}\|T_\alpha(t)\|_{L(X)},\quad \widetilde{M}_S=\sup_{0\leq t\leq T}Ce^{\omega t}(1+t^{1-\alpha}), $$ where $L(X)$ is the Banach space of bounded linear operators from $X$ into $X$ equipped with its natural topology. So we have $$ \|T_\alpha(t)\|_{L(X)}\leq \widetilde{M}_T,\quad \|S_\alpha(t)\|_{L(X)}\leq t^{\alpha-1}\widetilde{M}_S. $$ Let us consider the set of functions \begin{align*} PC(J,X) &= \{x:J\to X: x \in C((t_k,t_{k+1}],X),k=0,1,\dots p\text{ and there exist}\\ &\quad x(t_k^-)\text{ and } x(t_k^+),\; k=1,\dots,p\text{ with }x(t_k^-)=x(t_k)\}. \end{align*} Endowed with the norm $$ \|x\|_{PC}=\sup_{t\in J} \|x(t)\|_X, $$ the space $(PC(J,X),\|\cdot\|_{PC})$ is a Banach space. \begin{lemma}[\cite{chen}] \label{lem3.1} Consider the Cauchy problem \begin{gather*} D_t^\alpha x(t)+Ax(t)=f(t,x(t),x(a_1(t)),\dots,x(a_m(t))),\quad t>t_0,\;t_0\ge 0,\; 0<\alpha<1,\\ x(t_0)=x_0\in X, \end{gather*} if $f$ satisfies the uniform Holder condition with exponent $\beta\in(0,1]$ and $A$ is a sectorial operator, then the unique solution of this Cauchy problem is $$ x(t)=T_\alpha(t-t_0)x(t_0^+)+\int_{t_0}^tS_\alpha(t-\theta) f(\theta,x(\theta),x(a_1(\theta)),\dots,x(a_m(\theta)))d\theta, $$ where \begin{gather*} T_\alpha(t)= E_{\alpha,1}(-At^\alpha) =\frac{1}{2\pi i}\int_{\widehat{B}_r}e^{\lambda t} \frac{\lambda^{\alpha-1}}{\lambda^\alpha+A}d\lambda, \\ S_{\alpha}(t)= t^{\alpha-1}E_{\alpha,\alpha}(-At^\alpha) =\frac{1}{2\pi i}\int_{\widehat{B}_r}e^{\lambda t} \frac{1}{\lambda^\alpha+A}d\lambda, \end{gather*} where $\widehat{B}_r$ denotes the Bromwich path. $S_\alpha(t)$ is called the $\alpha-$resolvent family and $T_\alpha(t)$ is the solution operator, generated by $-A$. \end{lemma} \begin{proof} Let $t-t_0=s$, then $$ D_s^\alpha x(s+t_0)+Ax(s+t_0)=f(s+t_0,x(s+t_0),x(a_1(s+t_0)), \dots,x(a_m(s+t_0))), $$ for $s>0$. Now, applying the Laplace transform, we have \begin{equation} \label{d1} \begin{aligned} &\lambda^\alpha L\{x(s+t_0)\}-\lambda^{\alpha-1}x(t_0^+) +A L\{x(s+t_0)\}\\ &=L\{f(s+t_0,x(s+t_0),x(a_1(s+t_0)),\dots,x(a_m(s+t_0)))\}. \end{aligned} \end{equation} Since $(\lambda^\alpha I+A)^{-1}$ exists, that is $\lambda^\alpha\in\rho(A)$, from \eqref{d1}, we obtain \begin{align*} L\{x(s+t_0)\} &=\lambda^{\alpha-1}(\lambda^\alpha I+A)^{-1}x(t_0^+) +(\lambda^\alpha I+A)^{-1}\\ &\quad\times L\{f(s+t_0,x(s+t_0),x(a_1(s+t_0)), \dots,x(a_m(s+t_0)))\}. \end{align*} Therefore, by the inverse Laplace transform, we have \begin{align*} x(s+t_0)&=E_{\alpha,1}(-A s^\alpha)x(t_0^+)+\int_0^s(s-\tau)^{\alpha-1} E_{\alpha,\alpha}(-A(s-\tau)^\alpha)\\ &\quad\times f(\tau+t_0,x(\tau+t_0),x(a_1(\tau+t_0)), \dots,x(a_m(\tau+t_0)))d\tau. \end{align*} Let $s+t_0=t$, we obtain \begin{align*} x(t)&=E_{\alpha,1}(-A(t-t_0)^\alpha)x(t_0^+) +\int_0^{t-t_0}(t-t_0-\tau)^{\alpha-1}E_{\alpha, \alpha}(-A(t-t_0-\tau)^\alpha)\\ &\quad\times f(\tau+t_0,x(\tau+t_0),x(a_1(\tau+t_0)), \dots,x(a_m(\tau+t_0)))d\tau. \end{align*} This is the same as \begin{align*} x(t)&=E_{\alpha,1}(-A(t-t_0)^\alpha)x(t_0^+) +\int_{t_0}^t(t-\theta)^{\alpha-1}E_{\alpha,\alpha} (-A(t-\theta)^\alpha) \\ &\quad\times f(\theta,x(\theta),x(a_1(\theta)), \dots,x(a_m(\theta)))d\theta. \end{align*} Let $T_\alpha(t)=E_{\alpha,1}(-At^\alpha)$ and $S_\alpha(t)=t^{\alpha-1}E_{\alpha,\alpha}(-At^\alpha)$, then we have $$ x(t)=T_\alpha(t-t_0)x(t_0^+)+\int_{t_0}^tS_\alpha(t-\theta) f(\theta,x(\theta),x(a_1(\theta)),\dots,x(a_m(\theta)))d\theta. $$ This completes the proof of the Lemma. \end{proof} Now, we define the definition of mild solution of \eqref{ME}. \begin{definition}\label{mild} \rm A function $x\in$ $PC(J,X)$ solution of the fractional integral equation \begin{align*} x(t)=\begin{cases} T_\alpha(t)(x_0-g(x))\\ +\int_0^tS_\alpha(t-s)f(s,x(s),x(a_1(s)), \dots,x(a_m(s)))ds,& t\in[0,t_1]; \\[3pt] T_\alpha(t-t_1)[x(t_1^-)+I_1(x(t_1^-))]\\ +\int_{t_1}^t S_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_1,t_2];\\ \dots \\ T_\alpha(t-t_p)[x(t_p^-)+I_p(x(t_p^-))]\\ +\int_{t_p}^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_p,T]. \end{cases} \end{align*} will be called a mild solution of problem \eqref{ME}. From Lemma \ref{lem3.1} we can verify this definition. \end{definition} Now we introduce the following assumptions: \begin{itemize} \item[(H1)] There exists a constant $L_g>0$ such that $\|g(x)-g(y)\|_X\leq L_g\|x-y\|_X$. \item[(H2)] The nonlinear map $f:[0,T]\times X^{m+1}\to X$ is continuous and there exist a constant $L_f$ such that \begin{align*} &\|f(t,x_1,x_2,\dots,x_{m+1})- f(s,y_1,y_2,\dots,y_{m+1})\|_X\\ &\leq L_f\big[|t-s|+\sum_{i=1}^{m+1}\|x_i-y_i\|_X\big] \end{align*} for all $(x_1,\dots,x_{m+1})$ and $(y_1,\dots,y_{m+1})$ in $X^{m+1}$ and $t\in[0,T]$. \item[(H3)] The function $I_k:X\to X$ are continuous and there exists $L_k>0$ such that $\|I_k(x)-I_k(y)\|_X\leq L_k\|x-y\|_X,\quad x,y \in X,k=1,2,\dots,p,\;L=\max{\{L_k\}}>L_g$. \end{itemize} \begin{theorem} \label{thm3.3} Assume {\rm (H1)--(H3)} are satisfied and $$ \big[\widetilde{M}_T(1+L)+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}\big]<1. $$ Then impulsive problem \eqref{ME} has a unique mild solution $x\in PC(J,X)$. \end{theorem} \begin{proof} Define a mapping $N$ from $PC(J,X)$ into itself by \[ (N x)(t)=\begin{cases} T_\alpha(t)(x_0-g(x))\\ +\int_0^tS_\alpha(t-s)f(s,x(s),x(a_1(s)), \dots,x(a_m(s)))ds,& t\in[0,t_1]; \\[3pt] T_\alpha(t-t_1)[x(t_1^-)+I_1(x(t_1^-))]\\ +\int_{t_1}^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_1,t_2];\\ \dots \\ T_\alpha(t-t_p)[x(t_p^-)+I_p(x(t_p^-))]\\ +\int_{t_p}^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_p,T]. \end{cases} \] Now we show that $N$ is a contraction on $PC(J,X)$. We have \begin{align*} &\|Nx(t)-Ny(t)\|_X\\ &\leq\begin{cases} \|T_\alpha(t)\|_{L(X)}(\|g(x)-g(y)\|_X) +\int_0^t\|S_\alpha(t-s)\|_{L(X)} \\ \times \|f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))\\ -f(s,y(s),y(a_1(s)),\dots,y(a_m(s)))\|_Xds, & t\in[0,t_1]; \\[3pt] \|T_\alpha(t-t_1)\|_{L(X)}(\|x(t_1^-)-y(t_1^-)\|_X +\|I_1(x(t_1^-))-I_1(y(t_1^-))\|_X)\\ +\int_{t_1}^t\|S_\alpha(t-s)\|_{L(X)}\|f(s,x(s),x(a_1(s)), \dots,x(a_m(s)))\\ -f(s,y(s),y(a_1(s)),\dots,y(a_m(s)))\|_Xds, & t\in(t_1,t_2];\\ \dots \\ \|T_\alpha(t-t_p)\|_{L(X)}(\|x(t_p^-)-y(t_p^-)\|_X+\|I_p(x(t_p^-)) -I_p(y(t_p^-))\|_X)\\ +\int_{t_p}^t\|S_\alpha(t-s)\|_{L(X)}\|f(s,x(s),x(a_1(s)), \dots,x(a_m(s)))\\ -f(s,y(s),y(a_1(s)),\dots,y(a_m(s)))\|_X ds, & t\in(t_p,T]; \end{cases} \end{align*} Applying Assumptions (H1)--(H3), we obtain \[ \|Nx(t)-Ny(t)\|_X\leq \begin{cases} [\widetilde{M}_T[L_g+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}] \|x-y\|_{PC}, &t\in[0,t_1];\\ [\widetilde{M}_T(1+L_1)+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}]\|x-y\|_{PC}, &t\in(t_1,t_2]; \\ \dots \\ [\widetilde{M}_T(1+L_p)+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}]\|x-y\|_{PC}, &t\in(t_p,T]. \end{cases} \] Which implies that for $t\in[0,T]$, $$ \|Nx-Ny\|_{PC}\leq [\widetilde{M}_T(1+L)+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}]\|x-y\|_{PC}. $$ Since $[\widetilde{M}_T(1+L)+\widetilde{M}_SL_f(m+1) \frac{T^\alpha}{\alpha}]<1$, $N$ is a contraction. Therefore, $N$ has a unique fixed point by Banach contraction principle. This completes the proof of the theorem. \end{proof} Our second result is based on the following Krasnoselkii's fixed point theorem. \begin{theorem} \label{thm3.4} Let $B$ be a closed convex and nonempty subset of a Banach space $X$. Let $P$ and $Q$ be two operators such that: \begin{itemize} \item[(1)] $Px+Qy\in B$ whenever $x,y\in B$; \item[(2)] $P$ is compact and continuous; \item[(3)] $Q$ is a contraction mapping; \end{itemize} Then there exists $z\in B$ such that $z=Pz+Qz$. \end{theorem} Now, we make the following assumptions: \begin{itemize} \item[(H4)] $f\in C(J\times X^{m+1},X),\;g\in C(X,X)$, and $b_i\in C(J,J)$ $(i=1,\dots,m)$. Moreover, there are $C_i>0\;(i=1,2)$ such that $\|f(s,z_0,z_1,\dots,z_m)\|\leq C_1$ for $s\in J$, $z_i\in B_r\;(i=0,1,\dots,m)$ and $\|g(w)\|\leq C_2$ for $w\in X$. \item[(H5)] The function $I_k:X\to X$ are continuous and there exists $\rho>C_2$ such that $$ \rho=\max_{1\leq k\leq m,x\in B_r}\{\|I_k(x)\|_X\}. $$ \end{itemize} \begin{theorem}\label{thm3.5} Assume {\rm (H2), (H4), (H5)} are satisfied and $$ [\widetilde{M}_SL_f(m+1) {T^\alpha\over\alpha}]<1. $$ Then the impulsive problem \eqref{ME} has at least one mild solution on $J$. \end{theorem} \begin{proof} Choose $r\geq [\widetilde{M}_T(r+\rho) +\widetilde{M}_SC_1\frac{T^\alpha}{\alpha}]$ and consider $B_r=\{x\in PC(J,X):\|x\|_{PC}\leq r,\}$ then $B_r$ is a bounded, closed convex subset in $PC(J,X)$. Define on $B_r$ the operators $P$ and $Q$ by: \begin{gather*} (Px)(t)= \begin{cases} T_\alpha(t)(x_0-g(x)), & t\in[0,t_1];\\ T_\alpha(t-t_1)[x(t_1^-)+I_1(x(t_1^-))], & t\in(t_1,t_2];\\ \dots \\ T_\alpha(t-t_p)[x(t_p^-)+I_p(x(t_p^-))], & t\in(t_p,T], \end{cases} \\ (Qx)(t)= \begin{cases} \int_0^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in[0,t_1];\\ \int_{t_1}^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_1,t_2]; \\ \dots \\ \int_{t_p}^tS_\alpha(t-s)f(s,x(s),x(a_1(s)),\dots,x(a_m(s)))ds, &t\in(t_p,T]. \end{cases} \end{gather*} Now we present the proof in five steps: \textbf{Step 1.} We show that $Px+Qy\in B_r$ whenever $x,y\in B_r$. Let $x,y\in B_r$, then \begin{align*} &\|Px+Qy\|_{PC}\\ &\leq \begin{cases} \|T_\alpha(t)\|_{L(X)}(\|x_0\|_X+\|g(x)\|_X)\\ +\int_0^t\|S_\alpha(t-s)\|_{L(X)}\|f(s,y(s),y(a_1(s)), \dots,y(a_m(s)))\|_Xds, & t\in[0,t_1]; \\[3pt] \|T_\alpha(t-t_1)\|_{L(X)}[\|x(t_1^-)\|_X+\|I_1(x(t_1^-))\|_X]\\ +\int_{t_1}^t\|S_\alpha(t-s)\|_{L(X)}\|f(s,y(s),y(a_1(s)), \dots,y(a_m(s)))\|_Xds, & t\in(t_1,t_2]; \\ \dots \\ \|T_\alpha(t-t_p)\|_{L(X)}[\|x(t_p^-)\|_X+\|I_p(x(t_p^-))\|_X]\\ +\int_{t_p}^t\|S_\alpha(t-s)\|_{L(X)}\|f(s,y(s),y(a_1(s)), \dots,y(a_m(s)))\|_Xds, & t\in(t_p,T]. \end{cases} \\ &\leq \begin{cases} \widetilde{M}_T(r+C_2)+\widetilde{M}_SC_1\frac{T^\alpha}{\alpha}, & t\in[0,t_1];\\ \widetilde{M}_T(r+\rho)+\widetilde{M}_SC_1\frac{T^\alpha}{\alpha}, &t\in(t_1,t_2];\\ \dots \\ \widetilde{M}_T(r+\rho)+\widetilde{M}_SC_1\frac{T^\alpha}{\alpha}, & t\in(t_p,T]. \end{cases} \end{align*} Which implies \[ \|Px+Qy\|_{PC}\leq[\widetilde{M}_T(r+\rho) +\widetilde{M}_SC_1\frac{T^\alpha}{\alpha}] \leq r. \] \textbf{Step 2.} Continuity of $P$. For this purpose, let $\{x^n\}_{n=0}^\infty$ be a sequence in $B_r$ with $\lim x^n\to x$ in $B_r$. Then for every $t\in J$, we have \[ \|(Px^n)(t)-(Px)(t)\|_X \leq \begin{cases} \|T_\alpha(t)\|_{L(X)}\|g(x^n)-g(x)\|_X, & t\in[0,t_1];\\ \|T_\alpha(t-t_1)\|_{L(X)}[\|x^n(t_1^-)-x(t_1^-)\|_X\\ +\|I_1(x^n(t_1^-))-I_1x(t_1^-)\|_X], &t\in(t_1,t_2];\\ \dots \\ \|T_\alpha(t-t_p)\|_{L(X)}[\|x^n(t_p^-)-x(t_p^-)\|_X\\ +\|I_p(x^n(t_p^-))-I_px(t_p^-)\|_X], &t\in(t_p,T]. \end{cases} \] Since the functions $g$ and $I_k$, $k=1,\dots,p$ are continuous, $\lim_{n\to\infty}\|Px^n-Px\|_{PC}=0$ in $B_r$. This implies that the mapping $P$ is continuous on $B_r$. \textbf{Step 3.} $P$ maps bounded sets into bounded sets in $PC(J,X)$. So, let us prove that for any $r>0$ there exists a $\gamma>0$ such that for each $x\in B_r=\{x\in PC(J,X):\|x\|_{PC}\leq r\}$, we have $\|Px\|_{PC}\leq \gamma$. Indeed, we have for any $x\in B_r$, \begin{align*} \|Px(t)\|_X &\leq \begin{cases} \|T_\alpha(t)\|_{L(X)}(\|x_0\|_X+\|g(x)\|_X), & t\in[0,t_1]; \\ \|T_\alpha(t-t_1)\|_{L(X)}[\|x(t_1^-)\|_X+\|I_1(x(t_1^-))\|_X], & t\in(t_1,t_2]; \\ \dots \\ \|T_\alpha(t-t_p)\|_{L(X)}[\|x(t_p^-)\|_X+\|I_p(x(t_p^-))\|_X], &t\in(t_p,T]. \end{cases}\\ &\leq \begin{cases} \widetilde{M}_T(r+C_2), & t\in[0,t_1];\\ \widetilde{M}_T(r+\rho), & t\in(t_1,t_2];\\ \dots \\ \widetilde{M}_T(r+\rho), & t\in(t_p,T]. \end{cases} \end{align*} Which implies that $\|Px\|_{PC}\leq \widetilde{M}_T(r+\rho)=\gamma$. \textbf{Step 4.} We prove that $P(B_r)$ is equicontinuous with $B_r$. For $0\leq u