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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 222, pp. 1--42.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/222\hfil
 Existence, regularity and representation of solutions]
{Existence, regularity and representation of solutions of time fractional
wave equations}

\author[V. Keyantuo, C. Lizama, M. Warma \hfil EJDE-2017/222\hfilneg]
{Valentin Keyantuo, Carlos Lizama, Mahamadi Warma}

\address{Valentin Keyantuo \newline
University of Puerto Rico, Department of Mathematics,
Faculty of Natural Sciences, Rio Piedras Campus,  P.O. Box 70377,
San Juan, PR 00936-8377, USA}
\email{valentin.keyantuo1@upr.edu}

\address{Carlos Lizama \newline
Universidad de Santiago de Chile,
Departamento de Matem\'atica, Facultad de Ciencias,
Casilla 307-Correo 2, Santiago, Chile}
\email{carlos.lizama@usach.cl}

\address{Mahamadi Warma \newline
University of Puerto Rico,
Department of Mathematics,
Faculty of Natural Sciences,
Rio Piedras Campus,  P.O. Box 70377,
San Juan, PR 00936-8377, USA}
\email{mjwarma@gmail.com, mahamadi.warma1@upr.edu}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted October 26, 2016. Published September 18, 2017.}
\subjclass[2010]{47D06, 35K20, 35L20, 45N05}
\keywords{Fractional derivative; subordination principle; elliptic operator;
\hfill\break\indent integrated cosine family;
 Dirichlet, Neumann and Robin boundary conditions}

\begin{abstract}
 We study the  solvability of the fractional order inhomogeneous
 Cauchy problem
 $$
 \mathbb{D}_t^\alpha u(t)=Au(t)+f(t), \quad t>0,\;1<\alpha\le 2,
 $$
 where $A$ is a closed linear operator in
 some Banach space $X$ and $f:[0,\infty)\to X$ a given function.
 Operator families associated with this problem are defined and
 their regularity properties are investigated.  In the case where
 $A$ is a generator of a $\beta$-times integrated cosine family
 $(C_\beta(t))$, we derive explicit representations of mild and
 classical solutions of the above problem in terms of the
 integrated cosine family. We include applications to elliptic
 operators with Dirichlet, Neumann or Robin type boundary
 conditions on $L^p$-spaces and on the space  of continuous
 functions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

The classical wave equation provides the most important model for
the study of oscillation phenomena in physical sciences and
engineering. In the treatment of the evolutionary equation
\begin{equation}\label{Wave}
\frac{\partial^2 u(t,x)}{\partial t^2}=\Delta
u(t,x)+f(t,x), \quad t> 0,\, x\in\Omega,
\end{equation}
in function spaces over $\Omega$, where $\Omega\subset\mathbb{R}^N$ is an
open set, one needs initial conditions,
\[
u(0,x)=u_0(x), \quad\frac{\partial u(0,x)}{\partial t}=u_1(x),\quad
x\in\Omega;
\] and boundary conditions. Traditionally, Dirichlet and
Neumann boundary conditions are the most studied. The Robin type
boundary conditions,  $\nabla u\cdot \nu+\gamma u=g$ in
$\partial\Omega$ (where $\nu$ denotes the outer unit normal vector
at the boundary of the open set $\Omega$), have proven important due
to the fact that they arise naturally in heat conduction problems
as well as in physical Geodesy. Moreover, from the Robin boundary
conditions,  one can recover the Dirichlet and Neumann boundary
conditions (see e.g. \cite{AW1,AW2}). For more details and
applications we refer to \cite{AW1,AW2,Dan,Kel,Ote,War,War2} and
the references therein.

For many concrete problems it has been observed that equations of
fractional order in time provide a more suitable framework for their study.
Typical of this are phenomena with memory effects, anomalous diffusion,
problems in rheology, material science and several other areas.
We refer to the monographs \cite{Mi-Ro,Po99, Pr93} and the
papers \cite{Cl-Gr-Lo01, Cl-Lo-Si04, Ei-Ko04,Go-Ma97,Go-Ma00,Go-Lu-Ma99,Ma97,
Ni86, Za05} for more information.

We will investigate the linear inhomogeneous differential equation
of fractional order:
\begin{equation}\label{eq1.1}
\mathbb{D}_t^{\alpha}u(t)= Au(t) + f(t), \quad t>0, \; 1<\alpha \le 2,
\end{equation}
in which $\mathbb{D}_t^\alpha$ is the Caputo fractional
derivative. Here $X$ is a complex Banach space and $A$ is a closed
linear operator in $X$. The use of the Caputo fractional
derivative has the advantage (over, say, the Riemann-Liouville
fractional derivative) that the initial conditions are formulated
in terms of the values of the solution $u$ and its derivative at
$0$. These have physically significant interpretations in concrete
problems.


Our aim is to construct a basic theory for the solutions of this
equation along with applications to some partial differential
equations modeling phenomena from science and engineering. To
study the existence, uniqueness and regularity of the solutions of
Problem \eqref{eq1.1}, in general, one needs an operator family
associated with the problem \cite{LiSu, LiSu14}.  For example, the theory of cosine
families has been developed to deal with the case $\alpha=2$. In
case $A$ does not generate a  cosine family (if $\alpha=2$), the
concept of exponentially bounded $\beta$-times integrated cosine
families has been used in the treatment of Problem \eqref{eq1.1}.
In \cite{Ba01}, an operator family called $S_\alpha$ has been
introduced to deal with the fractional case, that is,
$1<\alpha\le 2$ and $\beta=0$.  Unfortunately, this theory does
not include the case of  exponentially bounded $\beta$-times
integrated  cosine families. Consequently,  the results obtained
in \cite{Ba01} cannot be applied to deal with the following
problem in $L^p(\Omega)$, $p\ne 2$, which is the fractional version
of \eqref{Wave}:
\begin{equation}\label{pp1}
\begin{gathered}
\mathbb D_t^\alpha u(t,x)-Au(t,x)=f(t,x),\quad
 t> 0,\; x\in\Omega,\; 1<\alpha\le 2,\\
 \frac{\partial u(t,z)}{\partial\nu_A} +\gamma(z)u(t, z)=0, \quad
 t> 0,\; z\in \partial\Omega,\\
 u(0,x)=u_0(x), \quad \frac{\partial u(0,x)}{\partial
t}=u_1(x),\quad  x\in\Omega.
\end{gathered}
\end{equation}
Here, $\Omega\subset\mathbb{R}^N$ ($N\ge 2$) is an open set with boundary
$\partial\Omega$, $A$ is a uniformly elliptic operator with bounded
measurable coefficients  formally given by
\begin{equation}\label{elip-dv}
Au=\sum_{j=1}^ND_j\Big(\sum_{i=1}^Na_{i,j}D_iu+b_ju\Big)
-\Big(\sum_{i=1}^Nc_iD_iu+du\Big)
\end{equation}
and
\[
\frac{\partial
u}{\partial\nu_A}=\sum_{j=1}^N\Big(\sum_{i=1}^Na_{ij}D_iu+b_ju\Big)\cdot\nu_j,
\]
where $\nu$ denotes the unit outer normal vector of $\Omega$ at
$\partial\Omega$ and $\gamma$ is a nonnegative measurable function in
$L^\infty(\partial\Omega)$ or $\gamma=\infty$.

In this paper, we introduce an appropriate operator family in a
general Banach space associated with Problem \eqref{eq1.1} that
will cover all the above mentioned cases. This family will be
called an $(\alpha,1)^\beta$-resolvent family $(\mathbb
S_\alpha^\beta(t))$ (see Definition \ref{def22} below) where
$1<\alpha\le 2$ and $\beta\ge 0$ is a real parameter associated
with the operator $A$. The case $\beta=0$ and $\alpha=2$
corresponds to the wave equation with $A$ generating a cosine
family. The family $\mathbb S_\alpha^0$ ($1<\alpha\le 2$)
corresponds to the family $S_\alpha$ introduced in the reference
\cite{Ba01} and mentioned above. The family $\mathbb
S_\alpha^\beta$, $\beta>0$ and  $\alpha=2$,  corresponds to the
theory of exponentially bounded $\beta$-times integrated  cosine
family. We use this framework to treat the homogeneous ($f=0$ in
\eqref{eq1.1}) as well as the inhomogeneous problems (under
suitable conditions on the function $f$ in \eqref{eq1.1}).  We
shall in fact consider the case where the operator $A$ is an
$L^p$-realization of a more general uniformly elliptic operator in
divergence form (as the one in \eqref{elip-dv})  with various
boundary conditions (Dirichlet, Neumann or Robin). We obtain a
representation of mild and classical solutions in terms of the
operator family $\mathbb S_\alpha^\beta$. Our results apply to the
situation where the closed linear operator $A$ satisfies the
following condition: There exist
 $\omega\ge 0$ and $\gamma\ge -1$ such that
 \begin{equation}\label{e-chi}
\| (\lambda^2-A)^{-1} \| \le M| \lambda|^\gamma,
\quad \operatorname{Re}(\lambda)>\omega.
\end{equation}
In fact, several operators of interest such as the Laplace
operator in $L^p(\mathbb{R}^N)$ for $N\ge 2$ and $p\ne 2$, which do not
generate cosine families are generators of integrated  cosine
families. See e.g. \cite[Chapter 8]{ABHN01} or \cite{ElKe,Hie}.
For the case of $L^p(\Omega)$, see e.g. \cite{KeWa,Ouh}. We refer to
the book of Brezis \cite[Section 10.3 and p.346]{Bre} for some
comments about the $L^p$-theory of the wave equation.

The paper is organized as follows. In Section \ref{pre}, we
present some preliminaries on fractional derivatives, the Wright
type functions and the Mittag-Leffler functions. In Section
\ref{sec3} we use the Laplace transform to motivate the
introduction of the operator family which will be used in the
sequel. Section \ref{sec33} is devoted to the definition and
several properties of the resolvent family $\mathbb
S_\alpha^\beta$. In the short Section \ref{sec4} we characterize
the resolvent family $\mathbb S_\alpha^\beta$ through the
regularized fractional Cauchy problem. The homogeneous
(fractional) abstract Cauchy problem is solved in Section
\ref{sec-ho-p} . The conditions on the initial data that ensure
solvability of the problem agree with the classical cases
$\alpha=2$. We take up the inhomogeneous (fractional) abstract
Cauchy problem in Section \ref{sec-inh-CP}. We are able to deal
satisfactorily with this problem under natural conditions on the
initial data and the inhomogeneity.  The results obtained in the
case $\alpha=2$ corresponding to integrated cosine families seem
to be new. In fact, we are able to deal with the full range
$1<\alpha\le 2$. In the final Section \ref{sec-app} we present
various examples of problems that can be handled with the results
obtained.

\section{Preliminaries}\label{pre}

The algebra of bounded linear operators on a Banach space $X$ will
be denoted by $\mathcal{L}(X),$  the resolvent set of a linear
operator $A$ by $\rho(A)$. We denote by $g_\alpha$ the function
$g_{\alpha}(t):= \frac{t^{\alpha-1}}{\Gamma(\alpha)}, t>0,\alpha
>0,$ where $\Gamma$ is the usual gamma function. It will be
convenient to write $g_0:= \delta_0$, the Dirac measure
concentrated at $0$. Note the semigroup property:
\begin{equation*}
g_{\alpha+ \beta } = g_{\alpha}*g_{\beta},\quad \alpha, \, \beta \ge 0.
\end{equation*}

The Riemann-Liouville fractional integral of order $\alpha>0$, of
a  locally integrable function $u:\;[0,\infty) \to X$ is given by:
\begin{equation*}%\label{eq-2.1a}
I_t^{\alpha}u(t): = (g_{\alpha}*u)(t) :=
\int_0^tg_{\alpha}(t-s)u(s)ds.
\end{equation*}
The Caputo fractional derivative of order $\alpha>0$ of a function
$u$ is defined by
\begin{equation*}
\mathbb{D}_t^\alpha u(t) := I_t^{m-\alpha} u^{(m)}(t) =
\int_0^tg_{m-\alpha}(t-s)u^{(m)}(s)ds
\end{equation*}
where $m:=\lceil\alpha\rceil$ is the smallest integer greatest
than or equal to $\alpha$, $u^{(m)}$ is the $m^{th}$-order
distributional derivative of $u(\cdot)$, under appropriate
assumptions. Then,  when $\alpha=n$ is a natural number, we get
$\mathbb{D}_t^n := \frac{d^n}{dt^n}$. In relation to the
Riemann-Liouville
 fractional derivative of order $\alpha$, namely $D_t^\alpha,$ we have:
\begin{equation}\label{RLC}
\mathbb{D}_t^\alpha
f(t)=D_t^\alpha\Big(f(t)-\sum_{k=0}^{m-1}f^{(k)}(0)g_{k+1}(t)\Big),
\quad t>0,
\end{equation}
where $m:=\lceil\alpha\rceil$ has been defined above, and for a
locally integrable function $u:\; [0,\infty) \to X$,
\[D_t^\alpha u(t):=\frac{d^m}{dt^m}\int_0^tg_{m-\alpha}(t-s)u(s)\,ds,\;\;\;t>0.\]

The Laplace transform of a locally integrable function $f:[0,\infty)\to X$
is defined by
\begin{align*}
 \mathcal{L}(f)(\lambda) :=  \widehat f(\lambda)
:=\int_0^{\infty} e^{-\lambda t} f(t) dt
=\lim_{R\to\infty}\int_0^{R} e^{-\lambda t} f(t)\,dt,
\end{align*}
provided the integral converges for some $\lambda\in\mathbb{C}$. If for
example $f$ is exponentially bounded, that is, there exist
$M\ge 0 $ and $\omega\ge 0$ such that $\| f(t)\|\le Me^{\omega t}$,
$t\ge 0$, then the integral converges absolutely  for
$\operatorname{Re}(\lambda) >\omega$ and defines an analytic function
there. The most general existence theorem for the Laplace
transform in the vector-valued setting is given by \cite[Theorem
1.4.3]{ABHN01}.

Regarding the fractional derivative, we have for $\alpha>0$ and
$m:=\lceil\alpha\rceil$, the following important properties:
\begin{gather}\label{LDC}
\widehat{ \mathbb{D}_t^\alpha  f}(\lambda) = \lambda^{\alpha}
\widehat f(\lambda) -
\sum_{k=0}^{m-1}\lambda^{\alpha-k-1}f^{(k)}(0), \\
%\label{LDRL}
\widehat{D_t^\alpha  f}(\lambda) = \lambda^{\alpha}
\widehat{f}(\lambda) - \sum_{k=0}^{m-1} (g_{m-\alpha}*f)^{(k)}(0)\lambda^{m-1-k}.
\nonumber
\end{gather}
The power function $\lambda^{\alpha}$ is uniquely defined as
$\lambda^{\alpha} = |\lambda|^{\alpha} e^{i\arg (\lambda)}, $
with $-\pi < \arg (\lambda) < \pi$.

Next, we recall some useful properties of convolutions that will
be frequently used throughout the paper. For every   $f\in
C([0,\infty);X)$, $k\in\mathbb{N}$, $\alpha\ge 0$ we have that for every
$t\ge 0$,
\begin{align}\label{conv1}
\frac{d^k}{dt^k}\left[(g_{k+\alpha}\ast
f)(t)\right]=(g_{\alpha}\ast f)(t).
\end{align}
Let $f\in  C([0,\infty);X)\cap C^1([0,\infty);X)$. Then for every $\alpha>0$ and $t\ge
0$,
\begin{align}\label{conv3}
\frac{d}{dt}\left[(g_\alpha\ast
f)(t)\right]=g_\alpha(t)f(0)+(g_\alpha\ast f')(t).
\end{align}
Let $k\in\mathbb{N}$. If $u\in C^{k-1}([0,\infty);X)$ and $v\in
C^k([0,\infty);X)$, then for every $t\ge 0$,
\begin{equation} \label{conv2}
\begin{aligned}
\frac{d^k}{dt^k}\left[(u\ast v)(t)\right]
=&\sum_{j=0}^{k-1}u^{(k-1-j)}(t)v^{(j)}(0)+(u\ast v^{(k)})(t) \\
=&\sum_{j=0}^{k-1}\frac{d^{k-1}}{dt^{k-1}}\left[(g_j\ast
u)(t)v^{(j)}(0)\right]+(u\ast v^{(k)})(t).
\end{aligned}
\end{equation}
The Mittag-Leffler function (see e.g.
\cite{Go-Ma00,Go-Lu-Ma99,Po99,SKM}) is defined as follows:
\begin{align}\label{mm1}
E_{\alpha, \beta}(z) :=
\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(\alpha n + \beta)} =
\frac{1}{2\pi i} \int_{Ha} e^{\mu}
\frac{\mu^{\alpha-\beta}}{\mu^{\alpha} - z}d\mu, \quad\alpha>0,
\;\beta \in\mathbb{C}, \; z \in \mathbb{C},
\end{align}
where $Ha$ is a Hankel path, i.e. a contour which starts and ends
at $-\infty$ and encircles the disc $ |\mu|\leq
|z|^{1/\alpha}$ counterclockwise. The Laplace transform
of the Mittag-Leffler function is given by (\cite{Po99}):
\begin{equation*}
\int_0^{\infty} e^{-\lambda t} t^{\alpha k + \beta-1}
E_{\alpha,\beta}^{(k)}(\pm \omega t^{\alpha})dt = \frac{k!
\lambda^{\alpha-\beta}}{(\lambda^{\alpha} \mp \omega)^{k+1}},
\quad \operatorname{Re}(\lambda)> |\omega|^{1/\alpha}.
\end{equation*}
Using this formula, we obtain for $0<\alpha\le 2$:
\begin{equation}\label{s-sca}
\mathbb{D}_t^\alpha E_{\alpha,1}(z t^{\alpha})= z E_{\alpha,1}(z
t^{\alpha}), \quad t>0, z \in \mathbb{C},
\end{equation}
that is, for every $z\in\mathbb{C}$, the function
$u(t):=E_{\alpha,1}(zt^\alpha)$ is a solution of the scalar valued
problem
\begin{align*}
\mathbb D_t^\alpha u(t)=zu(t),\;\;t> 0,\; 1<\alpha\le 2.
\end{align*}
In addition, one has the identity
\begin{equation*}%\label{eq2.6}
\frac{d}{dt}E_{\alpha,1}(zt^{\alpha})= zt^{\alpha
-1}E_{\alpha,\alpha}(zt^{\alpha}).
\end{equation*}
To see this, it is sufficient to write
\begin{align*}
\mathcal{L}\left(t^{\alpha-1}E_{\alpha,\alpha}(zt^{\alpha})\right)(\lambda)
=\frac{1}{\lambda^{\alpha} - z}
= \frac{1}{z} \big[ \lambda
\frac{\lambda^{\alpha-1}}{\lambda^{\alpha} - z} -1 \big],
\end{align*}
and invert the Laplace transform. Letting
$v(t):=E_{\alpha,1}(zt^\alpha)x$, $t>0$, $x\in X$, we have that
\begin{align}\label{new}
v(t)=g_1(t)x+ z(g_\alpha\ast v)(t).
\end{align}
By \cite[Formula (1.135)]{Po99} (or \cite[Formula (2.9)]{Ba01}),
if $\omega\ge 0$ is a real number, then there exist some constants
$C_1, C_2\ge 0$ such that
\begin{align}\label{Mit-E}
E_{\alpha,1}(\omega t^\alpha)\le
C_1e^{t\omega^{1/\alpha}}\;\text{ and }\;
E_{\alpha,\alpha}(\omega t^\alpha)\le
C_2e^{t\omega^{1/\alpha}},\quad t\ge 0,\;\alpha\in (0,2)
\end{align}
and the estimates in \eqref{Mit-E} are sharp. Recall the
definition of the Wright type function \cite[Formula
(28)]{Go-Lu-Ma99} (see also \cite{Po99,SKM,Wri}):
\begin{equation}\label{wright}
\Phi_{\alpha}(z):= \sum_{n=0}^{\infty} \frac{(-z)^n}{n!
\Gamma(-\alpha n +1 -\alpha)} = \frac{1}{2\pi i} \int_{\gamma}
\mu^{\alpha-1} e^{\mu - z \mu^{\alpha}} d\mu, \quad 0< \alpha <1,
\end{equation}
where $\gamma$ is a contour which starts and ends at $-\infty$ and
encircles the origin once counterclockwise.  This has sometimes
also been called the Mainardi function.  By \cite[p.14]{Ba01} or
\cite{Go-Lu-Ma99}, $\Phi_{\alpha}(t)$ is a probability density
function, that is,
\begin{align*}
\Phi_{\alpha}(t) \geq 0, \quad t>0; \quad \int_0^{\infty}
\Phi_{\alpha}(t)dt =1,
\end{align*}
and its Laplace transform is the Mittag-Leffler function in the
whole complex plane. We also have that
$\Phi_{\alpha}(0)=\frac{1}{\Gamma(1-\alpha)}$. Concerning the
Laplace transform of the Wright type functions, the following
identities hold:
\begin{gather}\label{eq2.3}
e^{-\lambda^{\alpha} s} = \mathcal{L}\Big(\alpha
\frac{s}{t^{\alpha+1}}
\Phi_{\alpha}(st^{-\alpha})\Big)(\lambda),\quad 0<\alpha<1,\\
\label{eq2.3-2}
\lambda^{\alpha-1}e^{-\lambda^{\alpha} s} = \mathcal{L}\Big(
\frac{1}{t^{\alpha}} \Phi_{\alpha}(st^{-\alpha})\Big)(\lambda),\quad 0<\alpha<1.
\end{gather}
See \cite[Formulas (40) and (42)]{Go-Lu-Ma99} and \cite[Formula
(3.10)]{Ba01}. We notice that the Laplace transform formula
\eqref{eq2.3} was formerly first given by Pollard  and Mikusinski
(see \cite{Go-Lu-Ma99} and references therein).

The following formula on  the moments of the Wright function will
be useful:
\begin{equation}\label{moment}
\int_0^\infty x^p
\Phi_\alpha(x)dx=\frac{\Gamma(p+1)}{\Gamma(\alpha p+1)},
\quad p+1>0,\; 0<\alpha<1.
\end{equation}
The preceding formula \eqref{moment} is derived from the
representation \eqref{wright} and can be found in
\cite{Go-Lu-Ma99}. For more details on the Wright type functions,
we refer to the papers \cite{Ba01,Go-Lu-Ma99,Ma97,Wri} and the
references therein. We note that the Wright functions have been
used by Bochner to
construct fractional powers of semigroup generators
(see e.g. \cite[Chapter IX]{Yo80}).

\section{Motivation} \label{sec3}

In this section we discuss heuristically the solvability of the
fractional order Cauchy problem \eqref{eq1.1}.
We proceed through the use of the Laplace transform and derive some
 representation formulas that will serve as motivation for the
theoretical framework of the subsequent sections.


Let $1<\alpha\le 2$ and suppose $u$ satisfies \eqref{eq1.1} and
that there exist some constants $M,\omega\ge 0$ such that
$\|(g_1\ast u)(t)\|\le Me^{\omega t}$, $t>0$. We rewrite the
fractional differential equation in integral form as:
 \begin{equation} \label{ee1}
u(t)=A(g_\alpha\ast u)(t)+(g_\alpha\ast
f)(t)+u(0)+tu^{\prime}(0),\quad t>0.
\end{equation}
Suppose also that $(g_1\ast f)(t)$ is exponentially bounded.
Taking the Laplace transform in both sides of \eqref{ee1} and
assuming that $\{\lambda^\alpha:\;
\operatorname{Re}(\lambda)>\omega\}\subset\rho(A)$ we have
\begin{equation} \label{ee2}
\widehat{u}(\lambda)=
\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}u(0)
+\lambda^{\alpha-2}(\lambda^\alpha-A)^{-1}u'(0)+
(\lambda^\alpha-A)^{-1}\widehat{f}(\lambda),
\end{equation}
for $\operatorname{Re}(\lambda)>\omega$.
Now we assume that $A$ is the generator of an exponentially
bounded $\beta$-times integrated cosine family $(C_\beta(t))$ on
$X$ for some $\beta\ge 0$, and denote by $(S_\beta(t))$ the
associated $(\beta+1)$-times integrated cosine family (or
$\beta$-times integrated sine family), namely,
$S_\beta(t)x=\int_0^t C_\beta(s)xds,\, t\ge 0$. Then by definition
there exist some constants $\omega, M\ge 0$ such that
$\|C_\beta(t)x\|\le Me^{\omega t}\|x\|$, $x\in X$, $t> 0$,
$\{\lambda^2\in\mathbb{C}: \operatorname{Re}(\lambda)>\omega\}\subset\rho(A)$
and
 \[
\lambda(\lambda^2-A)^{-1}x=\lambda^{\beta}\int_0^\infty
e^{-\lambda t}C_\beta(t)x\,dt
=\lambda^{\beta+1}\int_0^\infty e^{-\lambda t}S_\beta(t)xdt,
\]
for $\operatorname{Re}(\lambda)>\omega$, $x\in X$.
Substituting the above expression into \eqref{ee2} we arrive at
\begin{equation}  \label{ee333}
\begin{aligned}
\widehat{u}(\lambda)
=&\lambda^{\alpha-1}\lambda^{\frac{\alpha\beta}{2}
 -\frac{\alpha}{2}}\int_0^\infty e^{-\lambda^{\frac{\alpha}{2} }t} C_\beta(t)u(0)\,dt\\
& +\lambda^{\alpha-2}\lambda^{\frac{\alpha\beta}{2}
 -\frac{\alpha}{2}}\int_0^\infty e^{-\lambda^{\frac{\alpha}{2}}t}C_\beta(t)u'(0)dt
 +\lambda^{\frac{\alpha\beta}{2}-\frac{\alpha}{2}}\int_0^\infty e^{-\lambda^{\frac{\alpha}{2} }t}C_\beta(t)\widehat{f}(\lambda)dt \\
=&\lambda^{\frac{\alpha}{2}-1}\lambda^{\frac{\alpha\beta}{2}
}\int_0^\infty e^{-\lambda{s}}\int_0^\infty
\frac{\alpha t}{2 s^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})C_\beta(t)u(0)\,dsdt \\
&+\lambda^{\frac{\alpha}{2}-2}\lambda^{\frac{\alpha\beta}{2}
}\int_0^\infty e^{-\lambda{s}} \int_0^\infty
\frac{\alpha t}{2 s^{\frac{\alpha}{2}+1}}e^{-\lambda{s}}\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})C_\beta(t)u'(0)\,dsdt \\
&+ \lambda^{\frac{\alpha\beta}{2}-\frac{\alpha}{2}}\int_0^\infty
e^{-\lambda{s}}\int_0^\infty
\frac{\alpha t}{2 s^{\frac{\alpha}{2}+1}}e^{-\lambda{s}}\Phi_{\frac{\alpha}{2}}(st^{-\frac{\alpha}{2}})C_\beta(t)ds\widehat{f}(\lambda)dt \\
=&\lambda^{\frac{\alpha}{2}-1}\lambda^{\frac{\alpha\beta}{2}
}\int_0^\infty e^{-\lambda{s}}\int_0^\infty
\frac{\alpha t}{2 s^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})C_\beta(t)u(0)dtds \\
&+\lambda^{\frac{\alpha}{2}-2}\lambda^{\frac{\alpha\beta}{2}}\int_0^\infty
e^{-\lambda{s}} \int_0^\infty
\frac{\alpha t}{2 s^{\frac{\alpha}{2}+1}}e^{-\lambda{s}}\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})C_\beta(t)u'(0)dtds \\
&+ \lambda^{\frac{\alpha\beta}{2}-\frac{\alpha}{2}}\int_0^\infty
e^{-\lambda{s}}\int_0^\infty \frac{\alpha t}{2
s^{\frac{\alpha}{2}+1}}e^{-\lambda{s}}
\Phi_{\frac{\alpha}{2}}(st^{-\frac{\alpha}{2}})C_\beta(t)\widehat{f}(\lambda)dtds,
\end{aligned}
\end{equation}
where we have used the Laplace transform formula \eqref{eq2.3} and
Fubini's theorem.  Letting
\begin{align*}
 R_\alpha^\beta(t)x:=\int_0^\infty \frac{\alpha s}{2t^{\frac{\alpha}{2}+1}}
\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})C_\beta(s)xds,\quad t>0,
  \end{align*}
it follows from \eqref{ee333} that
\[
\widehat{u}(\lambda)=\lambda^{\frac{\alpha\beta}{2}}\widehat{(g_{1-\frac{\alpha}{2}}
 \ast
R_\alpha^\beta)}(\lambda)u(0)+\lambda^{\frac{\alpha\beta}{2}}\widehat{(g_{2-\frac{\alpha}{2}}\ast
R_\alpha^\beta)}u'(0)+\lambda^{\frac{\alpha\beta}{2}}\widehat{(g_{\frac{\alpha}{2}}\ast
R_\alpha^\beta\ast f)}(\lambda).
\]

If we use instead the associated "sine" function $(S_\beta(t))$,
we obtain the following representation
\begin{equation} \label{e-sine}
\begin{aligned}
\widehat{u}(\lambda)=&\lambda^{\frac{\alpha\beta}{2}}\lambda^{\alpha-1}\int_0^\infty e^{-\lambda s}\int_0^\infty\frac{\alpha t}{ 2s^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}(ts^{-\frac{\alpha}{2}})S_\beta(t)u(0)dtds \\
&+\lambda^{\frac{\alpha\beta}{2}}\lambda^{\alpha-2}\int_0^\infty e^{-\lambda s}\int_0^\infty\frac{\alpha t}{ 2s^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}(ts^{-{\frac{\alpha}{2}}})S_\beta(t)u'(0)dtds \\
&+\lambda^{\frac{\alpha\beta}{2}}\int_0^\infty e^{-\lambda
s}\int_0^\infty\frac{\alpha t}{
2s^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}(ts^{-{\frac{\alpha}{2}}})S_\beta(t)\widehat{f}(\lambda)\,dt\,ds.
\end{aligned}
\end{equation}
From this and using the uniqueness theorem for the Laplace
transform, we have the following:
\begin{gather*}
(g_{\frac{\alpha}{2}}\ast R_\alpha^\beta)(t)x=\int_0^\infty
\frac{\alpha s}{2t^{\frac{\alpha}{2}+1}}\Phi_{\frac{\alpha}{2}}
(st^{-\frac{\alpha}{2}})S_\beta(s)x\,ds, \quad t>0, \\
(g_{1-\frac{\alpha}{2}}\ast
R_\alpha^\beta)(t)x=D_t^{\alpha-1}(g_{\frac{\alpha}{2}}\ast
R_\alpha^\beta)(t)x,\quad  t>0, \\
(g_{2-\frac{\alpha}{2}}\ast R_\alpha^\beta)(t)x=(g_{2-\alpha}\ast
g_{\frac{\alpha}{2}}\ast R_\alpha^\beta)(t)x, \quad t>0.
 \end{gather*}

In the next section we will take inspiration from the above
heuristics to define and study the regularity properties of
resolvent families associated with Problem \eqref{eq1.1}. We will
also deal with the case when there is an underlying exponentially
bounded integrated  cosine family.



\section{Resolvent families and their properties}\label{sec33}

The following two definitions are motivated by the discussion in
Section \ref{sec3}.

\begin{definition}\label{def21} \rm
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$ and let $1<\alpha\le 2, \beta\ge 0$. We say
that $A$ is the generator of an $(\alpha,\alpha)^\beta$-resolvent
family if there exists  a strongly continuous function $\mathbb
P_{\alpha}^\beta:[0,\infty) \to \mathcal{L}(X)$  such that
$\|(g_1\ast \mathbb P_\alpha^\beta)(t)x\|\le Me^{\omega t}\|x||$,
$x\in X$, $t\ge 0$,  for some constants $M,\omega\ge 0$,
$\{\lambda^{\alpha} \, : \, \operatorname{Re}(\lambda)
> \omega \} \subset \rho(A)$,  and
\begin{align*}
(\lambda^{\alpha} -A)^{-1}x =
\lambda^{\frac{\alpha\beta}{2}}\int_0^{\infty} e^{-\lambda t}
\mathbb P_{\alpha}^\beta(t)x dt,  \quad  \operatorname{Re}(\lambda) >
\omega, \quad x \in X.
\end{align*}
In this case, $\mathbb P_{\alpha}^\beta$ is called the $(\alpha,
\alpha)^\beta$-resolvent family generated by $A$.
\end{definition}


\begin{definition}\label{def22} \rm
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$ and let $1<\alpha\le 2, \beta\ge 0$. We call
$A$ the generator of an $(\alpha,1)^\beta$-resolvent family if
there exists a strongly continuous function $\mathbb
S_{\alpha}^\beta:[0,\infty) \to \mathcal{L}(X)$  such that
$\|(g_1\ast\mathbb S_\alpha^\beta)(t)x\|\le Me^{\omega t}\|x||$,
$x\in X$, $t\ge 0$, for some $M,\omega\ge 0$, $\{\lambda^{\alpha}
\, : \,  \operatorname{Re}(\lambda)> \omega \} \subset \rho(A)$,  and
\begin{align*}
\lambda^{\alpha-1}(\lambda^{\alpha} -A)^{-1}x
=\lambda^{\frac{\alpha\beta}{2}} \int_0^{\infty} e^{-\lambda t}
\mathbb S_{\alpha}^\beta(t)x dt,  \quad  \operatorname{Re}(\lambda) >
\omega, \quad x \in X.
\end{align*}
In this case, $\mathbb S_{\alpha}^\beta$ is called the
$(\alpha,1)^\beta$-resolvent family generated by $A$.
\end{definition}

We will say that $\mathbb P_{\alpha}^\beta$ (resp. $\mathbb
S_{\alpha}^\beta$) is exponentially bounded if there exist some
constants $M,\omega\ge0$ such that $\|\mathbb
P_{\alpha}^\beta(t)\|\le Me^{\omega t},\;\forall t\ge 0$, (resp.
$\|\mathbb S_{\alpha}^\beta(t)\|\le Me^{\omega t},\;\forall t\ge
0$).

It follows from the uniqueness theorem for the Laplace transform
that an operator $A$ can generate at most one $(\alpha,1)^\beta$
(resp. $(\alpha,\alpha)^\beta$)-resolvent family for given
parameters $1<\alpha\le 2$ and $\beta\ge 0$.

We shall write $(\alpha,1) $ and $(\alpha,\alpha) $ for
$(\alpha,1)^0$ and $(\alpha,\alpha)^0$ respectively. Before we
give some  properties of the above resolvent families, we need the
following preliminary result.

\begin{lemma}\label{lem-ext-lap}
Let $f:\; [0,\infty)\to X$ be such that there exist some constants
$M\ge 0$ and $\omega\ge 0$ such that $\|(g_1\ast f)(t)\|\le
Me^{\omega t}$, $t>0$. Then for every $\alpha\ge 1$, there exist
some constants $M_1\ge 0$ and $\omega_1\ge 0$ such that
$\|(g_\alpha\ast f)(t)\|\le M_1e^{\omega_1 t}$, $t>0$.
\end{lemma}

\begin{proof}
Assume that $f$ satisfies the hypothesis of the lemma and let
$\alpha\ge 1$. We just have to consider the case $\alpha>1$. Then
for every $t\ge 0$,
\begin{align*}
\|(g_\alpha\ast f)(t)\|
&= \|(g_{\alpha-1}\ast g_1\ast f)(t)\|
 \le \int_0^tg_{\alpha-1}(s)Me^{\omega(t-s)}\,ds\\
&= Me^{\omega t}\int_0^t\frac{s^{\alpha-2}}{\Gamma(\alpha-1)}e^{-\omega s}\,ds\\
&\le Me^{\omega t}\frac{t^{\alpha-1}}{\Gamma(\alpha)}\le
M_1e^{\omega_1t},
\end{align*}
for some constants $M_1,\omega_1\ge 0$, and the proof is complete.
\end{proof}

\begin{remark}\label{rem33} \rm
 Let $A$ be a closed linear operator with domain $D(A)$
defined on a Banach space $X$ and let $1<\alpha\le 2, \beta\ge 0$.

(a)  Using Lemma \ref{lem-ext-lap} (this is used to show the exponential
boundedness) we have the following result. If $A$ generates an
 $(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$, then it
 generates an $(\alpha,\alpha)^\beta$-resolvent family $\mathbb P_\alpha^\beta$
 given by
\begin{equation}
\mathbb P_\alpha^\beta(t)x=(g_{\alpha-1}\ast \mathbb
S_\alpha^\beta)(t)x,\;\;\;t\ge 0,\;\;x\in X.
\end{equation}

(b) By the uniqueness theorem for  the Laplace transform, a
$(2,2)$-resolvent family corresponds to the concept of sine
family, while  a $(2,1)$-resolvent family corresponds to a cosine
family.  Furthermore,  a $(2,1)^\beta$-resolvent family
corresponds to the concept of exponentially bounded $\beta$-times
integrated cosine family. Likewise,   a $(2,2)^\beta$-resolvent
family represents an exponentially bounded $\beta$-times
integrated sine family. We refer to the monographs
\cite{ABHN01,Gol} and the corresponding references for a study of
the concepts of cosine and sine families and to \cite{ArKe} for an
overview on the theory of integrated cosine and sine families. A
systematic study in the fractional case is carried out in
\cite{Ba01} for the case $\beta=0$.
\end{remark}

Some properties of $(\mathbb P_{\alpha}^\beta(t))$ and $(\mathbb
S_{\alpha}^\beta(t))$ are included in the following lemmas. Their
proof uses techniques from the general theory of
$(a,k)$-regularized resolvent families \cite{Li00} (see also
\cite{Ar-Li88,Ba01}). It will be of crucial use in the
investigation of solutions of fractional order Cauchy problems in
Sections \ref{sec4}, \ref{sec-ho-p} and \ref{sec-inh-CP}. The
proof of the analogous results in the case of cosine families may
be found in \cite{ABHN01}. The corresponding result for the case
$0<\alpha\le 1$ is included in \cite{Ba01,KLW2013-2} for $\beta=0$
and in \cite{KLW-1} for $\beta\ge 0$. For the sake of completeness
we include the full proof.

 \begin{lemma}\label{le2.3-2}
 Let $A$ be a closed linear operator with domain $D(A)$ defined
on a Banach space $X$. Let $1<\alpha\le 2, \beta\ge 0$ and assume
that $A$ generates an $(\alpha,1)^\beta$-resolvent family $\mathbb
S_\alpha^\beta$.  Then the following properties hold:
 \begin{itemize}
\item[(a)] $\mathbb S_{\alpha}^\beta(t)D(A) \subset D(A)$ and
$A\mathbb S_{\alpha}^\beta(t)x =
\mathbb S_{\alpha}^\beta(t)Ax $ for all $x \in D(A), \, t \geq 0$.

\item[(b)] For all $x \in D(A)$,
\[
\mathbb S_{\alpha}^\beta(t)x = g_{\frac{\alpha\beta}{2}+1}(t)x +
\int_0^t g_{\alpha}(t-s)A\mathbb S_{\alpha}^\beta(s)xds, \quad t \geq
0.
\]

\item[(c)] For all $x \in X$, $(g_{\alpha}\ast \mathbb S_{\alpha}^\beta)(t)x \in
D(A)$ and
\[
\mathbb S_{\alpha}^\beta(t)x =
g_{\frac{\alpha\beta}{2}+1}(t)x +  A \int_0^t
g_{\alpha}(t-s)\mathbb S_{\alpha}^\beta(s)xds, \quad t \geq 0.
\]

\item[(d)] $\mathbb S_{\alpha}^\beta(0)=g_{\frac{\alpha\beta}{2}+1}(0)$.
Thus, $\mathbb S_{\alpha}^\beta(0)=I$ if $\beta=0$ and
$\mathbb S_{\alpha}^\beta(0)=0$ if $\beta>0$.
\end{itemize}
 \end{lemma}

\begin{proof}
Let $\omega$ be as in Definition \ref{def22}. Let
$\lambda,\mu>\omega$ and $x\in D(A)$. Then
$x=(I-\mu^{-\alpha}A)^{-1}y$ for some $y\in X$. Since
$(I-\mu^{-\alpha}A)^{-1}$ and $(I-\lambda^{-\alpha}A)^{-1}$ are
bounded and commute, and since the operator $A$ is closed, we
obtain from the definition of $\mathbb S_\alpha^\beta$ that,
\begin{align*}
\widehat{\mathbb S}_\alpha^\beta(\lambda)x
&=\int_0^\infty e^{-\lambda t} \mathbb S_\alpha^\beta(t)x\,dt \\
&=\widehat{\mathbb S}_\alpha^\beta(\lambda)(I-\mu^{-\alpha}A)^{-1}y\\
&=\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}\lambda^{-\alpha}(I-\lambda^{-\alpha}A)^{-1}(I-\mu^{-\alpha}A)^{-1}y\\
&=(I-\mu^{-\alpha}A)^{-1}\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}\lambda^{-\alpha}(I-\lambda^{-\alpha}A)^{-1}y\\
&=(I-\mu^{-\alpha}A)^{-1}\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}y\\
&=(I-\mu^{-\alpha}A)^{-1}\widehat{\mathbb S}_\alpha^\beta(\lambda)y\\
&=\int_0^\infty e^{-\lambda t} (I-\mu^{-\alpha}A)^{-1}\mathbb
S_\alpha^\beta(t)y\,dt.
\end{align*}
By the uniqueness theorem for the Laplace transform and by
continuity, we obtain
\begin{equation} \label{sc1-2}
\mathbb S_\alpha^\beta(t)x=(I-\mu^{-\alpha}A)^{-1}\mathbb
S_\alpha^\beta(t)y=(I-\mu^{-\alpha}A)^{-1}\mathbb
S_\alpha^\beta(t)(I-\mu^{-\alpha}A)x,\quad\forall t\ge 0.
\end{equation}
It follows from \eqref{sc1-2} that $\mathbb S_\alpha^\beta(t)x\in
D(A)$. Hence, $\mathbb S_\alpha^\beta(t)D(A)\subset D(A)$ for
every $t\ge 0$. It follows also from \eqref{sc1-2} that $A\mathbb
S_\alpha^\beta(t)x=\mathbb S_\alpha^\beta(t)Ax$ for all $x\in
D(A)$ and $t\ge 0$ and we have shown the assertion (a).

Next, let $x\in D(A)$. Using the convolution theorem, we get that
\begin{align*}
\int_0^\infty e^{-\lambda t}g_{\frac{\alpha\beta}{2}+1}(t)x\,dt
&=\lambda^{-\frac{\alpha\beta}{2}-1}x
 =\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}
 (I-\lambda^{-\alpha}A)x\\
&=\widehat{\mathbb S}_\alpha^\beta(\lambda)(I-\lambda^{-\alpha}A)x
 =\widehat{\mathbb S}_\alpha^\beta(\lambda)x-\lambda^{-\alpha}\widehat{S}_\alpha^\beta(\lambda)Ax\\
&=\int_0^\infty e^{-\lambda t}\Big[\mathbb
S_\alpha^\beta(t)x-\int_0^tg_{\alpha}(t-s)\mathbb
S_\alpha^\beta(s)Ax\,ds\Big]\,dt.
\end{align*}
By the uniqueness theorem for the Laplace transform we obtain the
assertion (b).

Next, let $\lambda\in \rho(A)$ be fixed, $x\in X$ and set
$y:=(\lambda-A)^{-1}x\in D(A)$. Let $z:=(g_\alpha\ast \mathbb
S_\alpha^\beta)(t)x$, $t\ge 0$. We have to show that $z\in D(A)$
and $Az=\mathbb
S_\alpha^\beta(t)x-g_{\frac{\alpha\beta}{2}+1}(t)x$. Using part
(b) we obtain that
\begin{align*}
z=&(\lambda-A)(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y=\lambda(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y-A(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y\\
=&\lambda(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y-(\mathbb
S_\alpha^\beta(t)y-g_{\frac{\alpha\beta}{2}+1}(t)y)\in D(A).
\end{align*}
Therefore,
\begin{align*}
Az=&\lambda A(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y-A\mathbb S_\alpha^\beta(t)y+g_{\frac{\alpha\beta}{2}+1}(t)Ay\\
=&\lambda (g_\alpha\ast A\mathbb S_\alpha^\beta)(t)y-\mathbb S_\alpha^\beta(t)Ay+g_{\frac{\alpha\beta}{2}+1}(t)(\lambda y-x)\\
=&\lambda (g_\alpha\ast A\mathbb S_\alpha^\beta)(t)y-\mathbb S_\alpha^\beta(t)(\lambda y-x)+g_{\frac{\alpha\beta}{2}+1}(t)(\lambda y-x)\\
=&\lambda\left[(g_\alpha\ast A\mathbb S_\alpha^\beta)(t)y-\mathbb S_\alpha^\beta(t) y+g_{\frac{\alpha\beta}{2}+1}(t) y\right]+\mathbb S_\alpha^\beta(t)x-g_{\frac{\alpha\beta}{2}+1}(t)x\\
=&\mathbb S_\alpha^\beta(t)x-g_{\frac{\alpha\beta}{2}+1}(t)x,
\end{align*}
and we have shown part (c).

Finally, it follows from the strong continuity of $\mathbb
S_\alpha^\beta(t)$ on $[0,\infty)$ and from the assertion (c) that
$\mathbb S_{\alpha}^\beta(0)x=g_{\frac{\alpha\beta}{2}+1}(0)x$ for
every $x\in X$. This implies the properties in part (d) and the
proof is finished.
\end{proof}

The corresponding result for the family $\mathbb P_\alpha^\beta$
is given in the following lemma. Its proof runs similar to the
proof of Lemma \ref{le2.3-2} and we shall omit it.

 \begin{lemma}\label{le2.3}
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$. Let $1<\alpha\le 2, \beta\ge 0$ and assume
that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent family
$\mathbb P_\alpha^\beta$.  Then the following properties hold.
\begin{itemize}

\item[(a)] $\mathbb P_{\alpha}^\beta(t)D(A) \subset D(A)$ and $A\mathbb P_{\alpha}^\beta(t)x =
\mathbb P_{\alpha}^\beta(t)Ax $ for all $x \in D(A), \, t\ge  0$.

\item[(b)] For all $x \in D(A)$, $\mathbb P_{\alpha}^\beta(t)x = g_{\alpha(\frac{\beta}{2}+1)}(t)x +
 \int_0^t g_{\alpha}(t-s)A\mathbb
P_{\alpha}^\beta(s)xds, \, t \ge 0$.

\item[(c)]  For all $x \in X$, $(g_{\alpha}\ast \mathbb P_{\alpha}^\beta)(t)x \in
D(A)$ and   $\mathbb P_{\alpha}^\beta(t)x =
g_{\alpha(\frac{\beta}{2}+1)}(t)x +  A \int_0^t
g_{\alpha}(t-s)\mathbb P_{\alpha}^\beta(s)xds, \, t \ge 0$.

\item[(d)]  $\mathbb P_{\alpha}(0)=g_{\alpha(\frac{\beta}{2}+1)}(0)=0$.
\end{itemize}
\end{lemma}

\begin{remark} \rm
 Let $A$ be a closed linear operator with domain $D(A)$
defined on a Banach space $X$. Let $1<\alpha\le 2$ and $\beta\ge
0$.
\begin{itemize}
\item[(i)] If $A$ generates an $(\alpha,1)^0=(\alpha,1)$-resolvent family
$\mathbb S_{\alpha}$, then it follows from Lemma \ref{le2.3-2}  (c) that
$D(A)$ is necessarily dense in $X$.

\item[(ii)] We notice that if $A$ generates an $(\alpha,1)^\beta$-resolvent
 family $\mathbb S_{\alpha}^\beta$ and $D(A)$ is dense in $X$ then this does
 not necessarily imply that $\beta=0$. Some examples will be given in
Section \ref{sec-app}.

\item[(iii)]  The examples presented below in Corollary \ref{cor-413}
show that in general ($\beta>0$) the domain of $A$ is not necessarily dense in $X$.

\end{itemize}
\end{remark}

 A family $\mathbb{S}(t)$ on $X$ is called non-degenerate if whenever we have
$\mathbb S(t)x=0$ for all $t\in [0,\tau]$ (for some fixed $\tau>0$),
then it follows that $x=0$. It follows from Lemma \ref{le2.3-2} and
Lemma \ref{le2.3} that the families $\mathbb S_\alpha^\beta$ and
$\mathbb P_\alpha^\beta$ are non-degenerate. We have the following
description of the generator $A$ of the resolvent family
$\mathbb S_\alpha^\beta$. We refer to \cite[Lemma 3.2.2]{ABHN01} for
related results in the case of integrated semigroups and
\cite[Proposition 3.14.5]{ABHN01} in the case of cosine families.
The corresponding result for the case $0<\alpha\le 1$ and $\beta\ge 0$
is contained in \cite[Proposition 6.8]{KLW-1} which was proved by using
the Laplace transform. Here, we provide an alternative proof which does not
 use the Laplace transform.

\begin{proposition}
Let $A$ be a closed linear operator on a Banach space $X$ with
domain $D(A)$. Let $1<\alpha\le 2$, $\beta\ge0$ and assume that
$A$ generates an $(\alpha,1)^\beta$- resolvent family
$\mathbb{S}_\alpha^\beta$. Then
\begin{align}\label{610}
A=\{(x,y)\in X\times X,\;\; \mathbb
S_\alpha^\beta(t)x=g_{\frac{\alpha\beta}{2}+1}(t)x+(g_\alpha\ast
\mathbb S_\alpha^\beta)(t)y,\;\forall t>0\}.
\end{align}
\end{proposition}

\begin{proof}
First we notice that since the  $(\alpha,1)^\beta$- resolvent
family $\mathbb S_\alpha^\beta$ is non-degenerate, the right hand
side of \eqref{610} defines a single-valued operator. Next, let
$x,y\in X$. We have to show that $x\in D(A)$ and $Ax=y$ if and
only if
\begin{equation}\label{611}
 \mathbb S_\alpha^\beta(t)x=g_{\frac{\alpha\beta}{2}+1}(t)x+(g_\alpha\ast \mathbb S_\alpha^\beta)(t)y,\;\forall t>0.
\end{equation}
Indeed, let $x\in D(A)$ and assume that $Ax=y$. Since $A$
generates an $(\alpha,1)^\beta$- resolvent family $\mathbb
S_\alpha^\beta$ and $Ax=y$, then \eqref{611} follows from Lemma
\ref{le2.3-2}. Conversely, let $x,y\in X$ and assume that
\eqref{611} holds. Let $\lambda\in\rho(A)$. It follows from
\eqref{611} and Lemma \ref{le2.3-2} that for all $t\in [0,\tau]$,
\begin{align*}
(\lambda-A)^{-1} (g_\alpha\ast \mathbb S_\alpha^\beta)(t)y&=(\lambda-A)^{-1}A(g_\alpha\ast \mathbb S_\alpha^\beta)(t)x\\
&=-(g_\alpha\ast \mathbb
S_\alpha^\beta)(t)x+\lambda(\lambda-A)^{-1}(g_\alpha\ast \mathbb
S_\alpha^\beta)(t)x.
\end{align*}
This implies
\begin{align*}
(g_\alpha\ast \mathbb S_\alpha^\beta)(t)\left[(\lambda-A)^{-1}
y+x-\lambda(\lambda-A)^{-1}x\right]=0.
\end{align*}
Since $\mathbb S_\alpha^\beta$ is non-degenerate, we have that
$(\lambda-A)^{-1} y+x-\lambda(\lambda-A)^{-1}x=0$ and this implies
that $x\in D(A)$ and $Ax=y$. The proof is finished.
\end{proof}


\begin{lemma}\label{lem-deri}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2, \beta\ge 0$. Assume that $A$ generates an
$(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$. Then
for every $x\in D(A)$ the mapping $t\mapsto \mathbb
S_\alpha^\beta(t)x$ is differentiable on $(0,\infty)$ and
\begin{equation}\label{eq-der}
(\mathbb
S_\alpha^\beta)'(t)x=g_{\frac{\alpha\beta}{2}}(t)x+\mathbb
P_\alpha^\beta(t)Ax,\quad t> 0.
\end{equation}
\end{lemma}

\begin{proof}
Let $x\in D(A)$. Then it is clear that the right-hand side of
\eqref{eq-der} belongs to $C((0,\infty),\mathcal L(X))$. Taking
the Laplace transform and using the fact that $\mathbb
S_\alpha^\beta(0)=0$, we get that for $\operatorname{Re}(\lambda)>\omega$
(where $\omega$ is the real number from the definition of $\mathbb
S_\alpha^\beta$ and $\mathbb P_\alpha^\beta$),
\begin{align*}
\widehat{(\mathbb
S_\alpha^\beta)'}(\lambda)x=\lambda\widehat{\mathbb
S_\alpha^\beta}(\lambda)x=\lambda\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}x=\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha}(\lambda^\alpha-A)^{-1}x.
\end{align*}
On the other hand, for $\operatorname{Re}(\lambda)>\omega$,
\begin{align*}
\widehat{g_{\frac{\alpha\beta}{2}}}(\lambda)x+\widehat{\mathbb
P_\alpha^\beta}(\lambda)Ax
&=\lambda^{-\frac{\alpha\beta}{2}}x+\lambda^{-\frac{\alpha\beta}{2}}
 (\lambda^\alpha-A)^{-1}Ax \\
&=\lambda^{-\frac{\alpha\beta}{2}}x-\lambda^{-\frac{\alpha\beta}{2}}(\lambda^\alpha-A)^{-1}(\lambda^\alpha-A-\lambda^\alpha)x\\
&=\lambda^{-\frac{\alpha\beta}{2}}x-\lambda^{-\frac{\alpha\beta}{2}}x
 +\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha}(\lambda^\alpha-A)^{-1}x \\
&=\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha}(\lambda^\alpha-A)^{-1}x.
\end{align*}
By the uniqueness theorem for the Laplace transform and continuity
of the right-hand side of \eqref{eq-der}, we conclude that  the
identity \eqref{eq-der} holds.
\end{proof}

Next, we give the principle of extrapolation of the families
$\mathbb S_\alpha^\beta$ and $\mathbb P_\alpha^\beta$ in terms of
the parameter $\beta$.

\begin{proposition}\label{prop-resol}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2,\beta\ge 0$. Then the following assertions hold.
\begin{itemize}
\item[(a)] If $A$ generates an $(\alpha,\alpha)^\beta$-resolvent family
 $\mathbb P_\alpha^\beta$, then it generates an $(\alpha,\alpha)^{\beta'}$-resolvent family $\mathbb P_\alpha^{\beta'}$ for every $\beta'\ge \beta$ and for every $x\in X$,
\begin{equation}\label{e1}
\mathbb
P_\alpha^{\beta'}(t)x=(g_{\alpha\frac{\beta'-\beta}{2}}\ast
\mathbb P_\alpha^\beta)(t)x,\;\;\;\forall t\ge 0.
\end{equation}

\item[(b)] If $A$ generates an $(\alpha,1)^\beta$-resolvent family
$\mathbb S_\alpha^\beta$, then it generates an $(\alpha,1)^{\beta'}$-resolvent family $\mathbb S_\alpha^{\beta'}$
for every $\beta'\ge \beta$ and for every $x\in X$,
\begin{equation}\label{e3}
\mathbb
S_\alpha^{\beta'}(t)x=(g_{\alpha\frac{\beta'-\beta}{2}}\ast
\mathbb S_\alpha^{\beta})(t)x,\;\;\forall t\ge 0.
\end{equation}
\end{itemize}
\end{proposition}

\begin{proof}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2,\beta\ge 0$.

(a) Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent
family $\mathbb P_\alpha^\beta$. Then,  by definition,  there
exists $\omega \geq0$  such that $\{\lambda^{\alpha} \, : \,
\operatorname{Re}(\lambda)
> \omega \} \subset \rho(A)$ and
\begin{align}\label{e-11}
(\lambda^{\alpha} -A)^{-1}x =
\lambda^{\frac{\alpha\beta}{2}}\int_0^{\infty} e^{-\lambda t}
\mathbb P_{\alpha}^\beta(t)x dt,  \quad  \operatorname{Re}(\lambda) >
\omega, \quad x \in X.
\end{align}
Let $\beta'\ge\beta$ and let $\mathbb
P_\alpha^{\beta'}$ be given in \eqref{e1}. Using Lemma
\ref{le2.3} we have that for every $x\in X$ and $t\ge 0$,
\begin{align*}
\mathbb P_\alpha^{\beta'}(t)x=&(g_{\alpha\frac{\beta'-\beta}{2}}\ast \mathbb P_\alpha^\beta)(t)x=(g_{\alpha\frac{\beta'-\beta}{2}}\ast g_{\alpha(\frac{\beta}{2}+1)})(t)x+ A\left(g_{\alpha\frac{\beta'-\beta}{2}}\ast g_\alpha\ast \mathbb P_\alpha^\beta\right)(t)x\\
=&g_{\alpha(\frac{\beta'}{2}+1)}(t)x
+A\left(g_{\alpha(\frac{\beta'-\beta}{2}+1)}\ast \mathbb
P_\alpha^\beta\right)(t)x.
\end{align*}
Hence,  $\mathbb P_\alpha^{\beta'}$ is strongly continuous
from $[0,\infty)$ into $\mathcal L(X)$. By \eqref{e1}, we have
that for every $x\in X$ and $t\ge 0$,
\begin{align*}
(g_1\ast\mathbb
P_\alpha^{\beta'})(t)x=(g_{\alpha\frac{\beta'-\beta}{2}+1}\ast
\mathbb P_\alpha^\beta)(t)x,
\end{align*}
and since by hypothesis $\|(g_1\ast \mathbb
P_\alpha^\beta)(t)x\|\le Me^{\omega t}\|x\|$, $x\in X$, $t\ge 0$,
for some constants $M,\omega\ge 0$,  it follows from Lemma
\ref{lem-ext-lap}  that there exist some constants
$M',\omega'\ge 0$ such that  $\|\mathbb (g_1\ast
P_{\alpha}^{\beta'})(t)x\|\le M' e^{\omega'
t}\|x\|$, $x\in X$, $t\ge 0$. Next, using \eqref{e-11}, we have
that for $\operatorname{Re}(\lambda) > \omega$, $x \in X$  and
$\beta'\ge \beta$,
\begin{align*}
(\lambda^{\alpha} -A)^{-1}x =&
\lambda^{\frac{\alpha\beta}{2}}\int_0^{\infty} e^{-\lambda t}
\mathbb P_{\alpha}^\beta(t)x dt=
\lambda^{\frac{\alpha\beta'}{2}}\lambda^{\alpha\frac{\beta-\beta'}{2}}\int_0^{\infty}
e^{-\lambda t}
\mathbb P_{\alpha}^\beta(t)x dt\\
=&\lambda^{\frac{\alpha\beta'}{2}}\int_0^\infty e^{-\lambda
s}g_{\alpha\frac{\beta-\beta'}{2}}(s)\,ds\int_0^{\infty}
e^{-\lambda t}
\mathbb P_{\alpha}^\beta(t)x dt\\
=&\lambda^{\frac{\alpha\beta'}{2}}\int_0^{\infty}
e^{-\lambda t} (g_{\alpha\frac{\beta-\beta'}{2}}\ast \mathbb
P_{\alpha}^\beta)(t)x dt=
\lambda^{\frac{\alpha\beta'}{2}}\int_0^{\infty} e^{-\lambda
t}
 \mathbb P_{\alpha}^{\beta'}(t)x dt.
\end{align*}
Hence, $A$ generates an $(\alpha,\alpha)^{\beta'}$-resolvent
family $\mathbb P_\alpha^{\beta'}$ given by \eqref{e1} and
we have shown the assertion (a).


(b) The proof of this part follows the lines of the proof of part
(a) where now we use Lemma \ref{le2.3-2}.
\end{proof}

The following example shows that a generation of an
$(\alpha,1)^{\beta}$ or $(\alpha,\alpha)^{\beta}$-resolvent family
does not imply a generation of an  $(\alpha',1)^{\beta}$ or
$(\alpha',\alpha')^{\beta}$-resolvent family for
$\alpha'>\alpha>1$. That is, an extrapolation property in
terms of the parameter $\alpha$ does not always hold.

\begin{example}\label{ex-ee} \rm
Let $1\le p<\infty$ and let $\Delta_p$ be the realization of
the Laplacian in $L^p(\mathbb{R}^N)$. It is well-known that $\Delta_p$
generates an analytic  $C_0$-semigroup of contractions of angle
$\pi/2$. Hence, for every $\varepsilon>0$,  there exists a
constant $C>0$ such that
\begin{equation}\label{e-csg}
\|(\lambda-\Delta_p)^{-1}\|\le\frac{C}{|\lambda|},\;\;\;\;
\lambda\in\Sigma_{\pi-\varepsilon}.
\end{equation}
where for $ 0<\gamma<\pi$, 
$\Sigma_{\gamma} := \{ z\in \mathbb{C} : 0<|arg(z)|<\gamma \}$.
Let $\theta\in [0,\pi)$ and let the operator $A_p$ on $L^p(\mathbb{R}^N)$
be given by $A_p:=e^{i\theta}\Delta_p$.  It follows from
\eqref{e-csg} that
\begin{equation} \label{dd}
\begin{aligned}
\|(\lambda-A_p)^{-1}\|
&=\|(\lambda-e^{i\theta}\Delta_p)^{-1}\|=\|(\lambda
e^{-i\theta}-\Delta_p)^{-1}\|\\
&\le\frac{C}{|\lambda|},\quad \lambda
e^{-i\theta}\in \Sigma_{\pi-\varepsilon}.
\end{aligned}
\end{equation}
Now, let $1<\alpha<2$. It follows from \eqref{dd} that, if
$\frac{\pi}{2}<\theta<\left(1-\frac{\alpha}{4}\right)\pi$,
then $\rho(A_p)\supset \Sigma_{\frac{\alpha\pi}{2} }$ and
\begin{align}\label{www}
\|(\lambda-A_p)^{-1}\|\le\frac{C}{|\lambda|},\quad \lambda \in
\Sigma_{\frac{\alpha\pi}{2}}.
\end{align}
By \cite[Proposition 3.8]{Ba01},  the estimate \eqref{www} implies
that $A_p$ generates an $(\alpha,1)=(\alpha,1)^0$-resolvent family
on $L^p(\mathbb{R}^N)$. Hence, by Proposition \ref{prop-resol} (c), $A_p$
generates an $(\alpha,1)^\beta$-resolvent family on $L^p(\mathbb{R}^N)$
for any $\beta\ge 0$. But one can verify by inspection of the
resolvent set of $A_p$ that it does not generate an
$(2,1)^\beta$-resolvent family, that is a $\beta$-times integrated
cosine family on $L^p(\mathbb{R}^N)$ for any $\beta\ge 0$.  However,
$A_p$ does generates a bounded analytic semigroup.
\end{example}


\begin{remark}\rm
In view of the asymptotic expansion of the Wright function
(see e.g. \cite{Go-Lu-Ma99,Wri}), for a locally  integrable
function $f:\; [0,\infty)\to X$ which is exponentially bounded at
infinity, and for any $0<\sigma<1$, the integral
$\int_0^\infty\Phi_{\sigma}(\tau)f(\tau)\,d\tau$ converges. This
property will be frequently used in the remainder of the article
without any mention.
\end{remark}

Concerning subordination of resolvent families we have the
following preliminary result.

\begin{lemma}\label{lem-ex-b}
Let $A$ be a closed linear operator on a Banach space $X$. Let
$1<\alpha\le 2$, $\beta\ge 0$. Then the following assertions hold.

(a)  Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent
family $\mathbb P_{\alpha}^\beta$. Let $1<\alpha'<\alpha$,
$\sigma:=\frac{\alpha'}{\alpha}$ and set
\begin{align}\label{E-CP11-2}
P(t)x:=\sigma t^{\sigma-1}\int_0^\infty s\Phi_\sigma(s)\mathbb
P_{\alpha }^\beta(st^\sigma)xds,\quad  t>0,\; x\in X.
\end{align}
Then $(g_1\ast P)(t)x$ is exponentially bounded. Moreover,
$(g_1\ast P)(t)x=\mathbb P(t)x$  where
\begin{equation} \label{E-CP11}
\mathbb P(t)x:=\int_0^\infty\frac{\sigma
s}{t^{\sigma+1}}\Phi_\sigma(st^{-\sigma})(g_{\frac{1}{\sigma}}
\ast\mathbb P_{\alpha }^\beta)(s)xds,\, \, t>0,\, x\in X.
\end{equation}

(b)  Assume that $A$ generates an $(\alpha,1)^\beta$-resolvent family
$\mathbb S_{\alpha}^\beta$. Let $1<\alpha'<\alpha$, $\sigma:=\frac{\alpha'}{\alpha}$
and set
\begin{equation} \label{E-CS11}
\mathbb
S(t)x:=\int_0^\infty\frac{1}{t^{\sigma}}\Phi_\alpha(st^{-\sigma})(g_{\frac{1}{\sigma}}\ast
\mathbb S_{\alpha}^\beta)(s)xds,\quad t> 0,\, x\in X.
\end{equation}
Then $\mathbb S$ is exponentially bounded. Moreover, $\mathbb
S(t)x=(g_1\ast S)(t)x$ where
\begin{equation} \label{E-CS11-2}
S(t)x=\int_0^\infty \Phi_\sigma(s)\mathbb
S_\alpha^\beta(st^\sigma)x\,ds, \quad \forall t\ge 0, \;x\in X.
\end{equation}
\end{lemma}

\begin{proof}
Let $A$, $\alpha$ and $\beta$ be as in the statement of the lemma.

(a) Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent
famlily $\mathbb P_{\alpha}^\beta$ and let
$1<\alpha'<\alpha$, $\sigma:=\frac{\alpha'}{\alpha}$
and $x\in X$.  Let $P(t)$ be given by \eqref{E-CP11-2}. By
hypothesis, there exist $M,\omega\ge 0$ such that $\|(g_1\ast
\mathbb P_{\alpha}^\beta)(t)x\|\le Me^{\omega t}\|x\|$ for every
$x\in X$, $t\ge 0$.  We show that there exist some constants
$M_1,\omega_1\ge 0$ such that for every $x\in X$, $\|(g_1\ast
P)(t)x\|\le M_1e^{\omega_1 t}\|x\|$, $t\ge 0$. Using
\eqref{E-CP11-2}, Fubini's theorem, \eqref{moment}, \eqref{mm1}
and \eqref{Mit-E}, we get  that for every $t\ge 0$ and $x\in X$,
\begin{align*}
\|\int_0^t P(\tau)x\,d\tau\|
\le &\int_0^\infty s\Phi_{\sigma}(s)\|\int_0^t \sigma \tau^{\sigma-1}\mathbb
P_\alpha^\beta(s \tau^\sigma)x\,d\tau\|\,ds\\
=& \int_0^\infty \Phi_{\sigma}(s)\|\int_0^{st^\sigma} \mathbb P_\alpha^\beta(\tau)x\,d\tau\|\,ds\\
\le& M\|x\| \int_0^\infty \Phi_{\sigma}(s)e^{\omega st^\sigma}\,ds \\
= & M\|x\|\sum_{n=0}^\infty\frac{(\omega t^\sigma)^n}{n!} \int_0^\infty \Phi_{\sigma}(s)s^n\,ds\\
\le &M\|x\|\sum_{n=0}^\infty\frac{(\omega t^\sigma)^n}{n!}
\frac{\Gamma(n+1)}{\Gamma(\sigma n+1)}\\
= & M\|x\|\sum_{n=0}^\infty\frac{(\omega t^\sigma)^n}{\Gamma(\sigma n+1)}
=M\|x\|E_{\sigma,1}(\omega t^\sigma)\\
\le& M_1e^{t\omega^{\frac{1}{\sigma}}}\|x\|,
\end{align*}
for some constant $M_1\ge 0$. Taking the Laplace transform of
\eqref{E-CP11} by using \eqref{eq2.3} and Fubini's theorem, we
have that for $\operatorname{Re}>\omega$ and $x\in X$,
\begin{align*}
\int_0^\infty e^{-\lambda t}\mathbb P(t)x\,dt=&\int_0^\infty e^{-\lambda t}\int_0^\infty\frac{\sigma s}{t^{\sigma+1}}\Phi_\sigma(st^{-\sigma})(g_{\frac{1}{\sigma}} \ast\mathbb P_{\alpha }^\beta)(s)xds\,dt\\
=&\int_0^\infty e^{-\lambda^\sigma s}(g_{\frac{1}{\sigma}}
\ast\mathbb P_{\alpha }^\beta)(s)xds
=\lambda^{-1}\lambda^{-\frac{\alpha'\beta}{2}}(\lambda^{\alpha'}-A)^{-1}x.
\end{align*}
Similarly, we have that for $\operatorname{Re}>\omega$ and $x\in X$,
\begin{align*}
\int_0^\infty e^{-\lambda t}(g_1\ast P)(t)x\,dt
=&\lambda^{-1}\int_0^\infty e^{-\lambda t}P(t)x\,dt \\
=&\lambda^{-1}\int_0^\infty e^{-\lambda t}\sigma t^{\sigma-1}
 \int_0^\infty s\Phi_\sigma(s)\mathbb P_{\alpha }^\beta(st^\sigma)x\,ds\,dt\\
=&\lambda^{-1}\int_0^\infty \mathbb P_\alpha^\beta(\tau)x
 \int_0^\infty e^{-\lambda t}\frac{\sigma\tau}{t^{\sigma+1}}
 \Phi_\sigma(\tau t^{-\sigma})\,dt\,d\tau \\
=&\lambda^{-1}\int_0^\infty e^{-\tau\lambda^\sigma}
 \mathbb P_\alpha^\beta(\tau)x\,d\tau\\
=&\lambda^{-1}\lambda^{-\frac{\alpha'\beta}{2}}(\lambda^{\alpha'}-A)^{-1}x.
\end{align*}
By the uniqueness theorem for the Laplace transform and by
continuity, we have that $(g_1\ast P)(t)x=\mathbb P(t)x$ for all
$t\ge 0$ and $x\in X$ and this completes the proof of part (a).

(b) Assume that $A$ generates an $(\alpha,1)^\beta$-resolvent
family $\mathbb S_{\alpha}^\beta$ and let
$1<\alpha'<\alpha$, $\sigma:=\frac{\alpha'}{\alpha}$
and $x\in X$. Then there exist some constants $M,\omega\ge 0$ such
that $\|(g_1\ast \mathbb S_\alpha^\beta)(t)x\|\le Me^{\omega
t}\|x\|$, $t\ge 0$. Since $\frac{1}{\sigma}>1$, it follows from
Lemma \ref{lem-ext-lap} that there exist some constants
$M_1,\omega_1\ge 0$ such that for every $t\ge 0$ and $x\in X$,
\begin{align}\label{es-ex-2}
\|(g_{\frac{1}{\sigma}}\ast \mathbb S_\alpha^\beta)(t)x\|\le
M_1e^{\omega_1 t}\|x\|.
\end{align}
 Using \eqref{E-CS11}, \eqref{moment}, \eqref{es-ex-2}, \eqref{mm1} and \eqref{Mit-E}, we have that for every $x\in X$, $t>0$,
\begin{align*}
\|S(t)x\|
\le &M_1\|x\|\int_0^\infty \frac{1}{t^{\sigma}}\Phi_\sigma(st^{-\sigma})
 e^{\omega_1 s}\,ds=M_1\|x\|\int_0^\infty \Phi_\sigma(\tau) e^{\omega_1 \tau t^\sigma}\,d\tau\\
\le &M_1\|x\|\sum_{n=0}^\infty\frac{(\omega_1 t^\sigma)^n}{n!}
 \int_0^\infty\Phi_\sigma(\tau) \tau^n\,d\tau \\
= &M_1\|x\|\sum_{n=0}^\infty\frac{(\omega_1 t^\sigma)^n}{n!}\frac{\Gamma(n+1)}{\Gamma(\sigma n+1)}\\
\le &M_1\|x\|\sum_{n=0}^\infty\frac{(\omega_1
t^{\sigma})^n}{\Gamma(\sigma n+1)}=M_1E_{\sigma,1}(\omega_1
t^\sigma)\|x\|\\
\le &Me^{t\omega_1^{\frac{1}{\sigma}}}\|x\|,
\end{align*}
for some constant $M\ge 0$. This completes the proof.
\end{proof}


Next, we present the principle of subordination of the families
$\mathbb S_\alpha^\beta$ and $\mathbb P_\alpha^\beta$ in terms of
the parameter $\alpha$.

\begin{theorem}\label{th-resol-22}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2,\beta\ge 0$. Then the following assertions hold.
\begin{itemize}
\item[(a)] If $A$ generates an $(\alpha,\alpha)^\beta$-resolvent family $\mathbb P_\alpha^\beta$, then it generates an $(\alpha',\alpha')^{\beta }$-resolvent family $\mathbb P_{\alpha'}^\beta$ for every $1<\alpha'<\alpha$ and
\begin{equation}\label{e2}
\mathbb P_{\alpha'}^\beta(t)x=\sigma
t^{\sigma-1}\int_0^\infty s\Phi_{\sigma}(s)\mathbb
P_\alpha^\beta(s t^\sigma)x\,ds,\;\;\forall t> 0,\;x\in
X,
\end{equation}
 where $\sigma:=\frac{\alpha'}{\alpha}$.

\item[(b)] If $A$ generates an $(\alpha ,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$, then it generates an $(\alpha',1)^{\beta }$-resolvent family $\mathbb S_{\alpha'}^{\beta}$
for every $1<\alpha'<\alpha$ and
\begin{equation}\label{ee3}
\mathbb S_{\alpha'}^{\beta}(t)x=\int_0^\infty
\Phi_\sigma(s)\mathbb S_\alpha^\beta(st^\sigma)x\,ds,
\;\;\forall t\ge 0, \;x\in X,\;\text{ where }\;\;
\sigma:=\frac{\alpha'}{\alpha}.
\end{equation}
\end{itemize}
\end{theorem}

\begin{proof}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2,\beta\ge 0$.

(a) Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent
family $\mathbb P_\alpha^\beta$. Let $1<\alpha'<\alpha$ and
let $\mathbb P_{\alpha'}^\beta$ be given by \eqref{e2}. Then
it is clear that $\mathbb P_{\alpha'}^\beta$ is strongly
continuous from $(0,\infty)$ into $\mathcal L(X)$. We show that
$\mathbb P_{\alpha'}^\beta(t)$ is strongly continuous at
$0$. Since $\mathbb P_{\alpha}^\beta(t)\simeq
g_{\alpha(\frac{\beta}{2}+1)}(t)=\frac{t^{\alpha(\frac{\beta}{2}+1)-1}}{\Gamma(\alpha(\frac{\beta}{2})+1}$
as $t\to 0$, we get from \eqref{e2} that
\begin{align*}
\mathbb P_{\alpha'}^\beta(t)\simeq
t^{\frac{\alpha'}{\alpha}-1}
t^{\frac{\alpha'}{\alpha}\alpha(\frac{\beta}{2}+1)-\frac{\alpha'}{\alpha}}
=t^{\alpha'(\frac{\beta}{2}+1)-1}\;\;\;\text{ as }\, t\to 0.
\end{align*}
We have shown that $\mathbb P_{\alpha'}^\beta(t)$ is
strongly continuous at $0$.  By Lemma \ref{lem-ex-b}(a),  there
exist some constants $M_1,\omega_1\ge 0$ such that $\|(g_1\ast
\mathbb P_{\alpha'}^\beta)(t)x\|\le M_1e^{\omega_1 t}$,
$x\in X$, $t\ge 0$. Now, it follows from \eqref{e-11} and
\eqref{eq2.3} that $\{\lambda^{\alpha'}:\;
\operatorname{Re}(\lambda)>\omega\}\subset \rho(A)$ and for
$\operatorname{Re}(\lambda) > \omega$, $x \in X$,
\begin{align*}
(\lambda^{\alpha'} -A)^{-1}x
=& \lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty}
e^{-\lambda^{\sigma} t} \mathbb P_{\alpha}^\beta(t)x dt \\
=& \lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty} e^{-\lambda
t}\int_0^\infty\sigma
\frac{s}{t^{\sigma+1}}\Phi_{\sigma}(st^{-\sigma})
\mathbb P_{\alpha}^\beta(s)x ds\,dt\\
=&\lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty}
e^{-\lambda t} \sigma t^{\sigma-1}\int_0^\infty s \Phi_{\sigma}(s)
\mathbb P_{\alpha}^\beta(st^{\sigma})x
ds\,dt\\
=&\lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty}
e^{-\lambda t}\mathbb P_{\alpha'}^\beta(t)x\,dt.
\end{align*}
Hence, $A$ generates an
$(\alpha',\alpha')^{\beta}$-resolvent family $\mathbb
P_{\alpha'}^{\beta'}$ given by \eqref{e2} and we have
shown part (a).

(b) Now assume that $A$ generates an $(\alpha,1)^\beta$-resolvent
family $\mathbb S_\alpha^\beta$. Then by definition, there exists
$\omega\ge 0$ such that
$\{\lambda^\alpha:\;\operatorname{Re}(\lambda)>\omega\}\subset\rho(A)$ and
\begin{equation}\label{de-resol}
\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}x=\lambda^{\frac{\alpha\beta}{2}}
\int_0^\infty e^{-\lambda t}\mathbb
S_\alpha^\beta(t)x\,dt,\;\operatorname{Re}(\lambda)>\omega,\;\forall
x\in X.
\end{equation}
Let $1<\alpha'<\alpha$ and let $\mathbb
S_{\alpha'}^\beta$ be given by \eqref{ee3}. Then it is clear
that $\mathbb S_{\alpha'}^\beta$ is strongly continuous from
$[0,\infty)$ into $\mathcal L(X)$.  By Lemma \ref{lem-ex-b}(b),
there exist some constants $M_1,\omega_1\ge 0$ such that for every
$x\in X$, $\|(g_1\ast \mathbb S_{\alpha'}^\beta)(t)x\|\le
M_1e^{\omega_1 t}\|x\|$, $t\ge 0$. It follows from
\eqref{de-resol} and \eqref{eq2.3-2} that
$\{\lambda^{\alpha'}:\; \operatorname{Re}(\lambda)>\omega\}\subset
\rho(A)$ and for $\operatorname{Re}(\lambda) > \omega$, $x \in X$,
\begin{align*}
\lambda^{\alpha'-1}(\lambda^{\alpha'} -A)^{-1}x
=& \lambda^{\frac{\alpha'\beta}{2}}\lambda^{\sigma-1}\int_0^{\infty}
e^{-\lambda^{\sigma} t} \mathbb S_{\alpha}^\beta(t)x dt \\
=& \lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty} e^{-\lambda
t}\int_0^\infty \frac{1}{t^{\sigma}}\Phi_{\sigma}(st^{-\sigma})
\mathbb S_{\alpha}^\beta(s)x ds\,dt\\
=&\lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty}
e^{-\lambda t}\int_0^\infty  \Phi_{\sigma}(s) \mathbb
S_{\alpha}^\beta(st^{\sigma})x\,ds\,dt\\
=&\lambda^{\frac{\alpha'\beta}{2}}\int_0^{\infty}
e^{-\lambda t}\mathbb S_{\alpha'}^\beta(t)x\,dt.
\end{align*}
Hence, $A$ generates an $(\alpha',1)^{\beta}$-resolvent
family $\mathbb S_{\alpha'}^{\beta'}$ given by
\eqref{ee3}. The proof of the theorem is finished.
\end{proof}

We have the following result as a corollary of the preceding
theorem.

\begin{corollary}\label{cor-413}
Let $1<\alpha\le 2$, $\beta\ge 0$ and let $A$ be a closed linear
operator on a Banach space $X$. If $A$ generates an exponentially
bounded $\beta-$times integrated cosine family $(C_\beta(t))$,
then $A$ generates an exponentially bounded
$(\alpha,1)^\beta$-resolvent family $(\mathbb
S_{\alpha}^\beta(t))$ given by
\begin{equation}\label{eq4.1a}
\mathbb S_{\alpha}^\beta(t)x = \int_0^{\infty}
t^{-\frac{\alpha}{2}}
\Phi_{\frac{\alpha}{2}}(st^{-\frac{\alpha}{2}}) C_\beta(s)x ds =
\int_0^{\infty}  \Phi_{\frac{\alpha}{2}}(\tau) C_\beta(\tau
t^{\frac{\alpha}{2}})x d\tau,
\end{equation}
for $t> 0$, $x \in X$.
In particular, it follows from the first representation formula in
\eqref{eq4.1a} that $(\mathbb S_{\alpha}^\beta(t))$ is analytic
for $t>0$, and, from the second one, that $\mathbb
S_{\alpha}^\beta(0)=C_\beta(0)$.

Let $(\mathbb P_{\alpha}^\beta(t))$ be the associated
$(\alpha,\alpha)^\beta$-resolvent family generated by $A$ which
exists by Remark \ref{rem33} (b). Then for every $x\in X$,
\begin{equation} \label{eq4.3a}
\begin{aligned}
\mathbb P_{\alpha}^\beta(t)x
& =  \frac{\alpha}{2}
\int_0^{\infty} \frac{s}{t^{\frac{\alpha}{2} +1}}
\Phi_{\frac{\alpha}{2}}(st^{-\frac{\alpha}{2}}) C_\beta(s)x ds \\
&=  \frac{\alpha}{2}  \int_0^{\infty}
\frac{\tau}{t^{1-\frac{\alpha}{2}}} \Phi_{\frac{\alpha}{2}}(\tau)
C_\beta(\tau t^{\frac{\alpha}{2}})x d\tau,
\end{aligned}
\end{equation}
for $t>0$.
\end{corollary}

\begin{proof}
Let $\alpha$, $\beta$ and $A$ be as in the statement of the
theorem. The fact that $A$ generates an
$(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$ and
an $(\alpha,\alpha)^\beta$-resolvent family $\mathbb
P_\alpha^\beta$ is a direct consequence of Theorem
\ref{th-resol-22} since by hypothesis $A$ generates a
$\beta$-times integrated cosine family, that is a
$(2,1)^\beta$-resolvent family, and a $\beta$-times integrated
sine family, that is a $(2,2)^\beta$-resolvent family. The
formulas \eqref{eq4.1a} and \eqref{eq4.3a} are the corresponding
formulas \eqref{ee3} and \eqref{e2}, respectively, in Theorem
\ref{th-resol-22}. It remains to show that $\mathbb
S_\alpha^\beta$ and $\mathbb P_\alpha^\beta$ are exponentially
bounded. By hypothesis,  $(C_\beta(t))$ is exponentially bounded,
that is, there exist some constants $M,\omega\ge 0$ such that
$\|C_\beta(t)x\|\le Me^{\omega t}\|x||$ for all $t\ge 0$, $x\in
X$. Using \eqref{eq4.1a}, \eqref{moment}, \eqref{mm1} and
\eqref{Mit-E}, we have that for every $x\in X$, $t\ge 0$,
\begin{align*}
\|\mathbb S_{\alpha}^\beta(t)x \|\le &\int_0^{\infty}
\Phi_{\frac{\alpha}{2}}(\tau) \|C_\beta(\tau
t^{\frac{\alpha}{2}})x \|d\tau\le M\|x\|\int_0^{\infty}
\Phi_{\frac{\alpha}{2}}(\tau)e^{\omega \tau
t^{\frac{\alpha}{2}}}\,d\tau\\
\le &M\|x\|\sum_{n=0}^\infty\frac{(\omega
t^{\frac{\alpha}{2}})^n}{n!}\int_0^\infty
\Phi_{\frac{\alpha}{2}}(\tau)\tau^n\,d\tau=
M\|x\|\sum_{n=0}^\infty\frac{(\omega t^{\frac{\alpha}{2}})^n}{n!}\frac{\Gamma(n+1)}{\Gamma(\frac{\alpha}{2}n+1)}\\
\le &M\|x\|\sum_{n=0}^\infty\frac{(\omega
t^{\frac{\alpha}{2}})^n}{\Gamma(\frac{\alpha}{2}n+1)}=M\|x\|E_{\frac{\alpha}{2},1}(\omega
t^{\frac{\alpha}{2}})\le M_1e^{t\omega^{\frac{2}{\alpha}}}\|x\|,
\end{align*}
for some constant $M_1\ge 0$ and we have shown that $\mathbb
S_\alpha^\beta$ is exponentially bounded.

We note that $\mathbb P_\alpha^\beta$ is bounded in a neighborhood
of $t=0^+$  by strong continuity on $[0,\infty)$. We show that
 $\mathbb P_\alpha^\beta$ is exponentially bounded away from
$0$. Indeed, using \eqref{eq4.3a}, \eqref{moment}, \eqref{mm1}
and \eqref{Mit-E}, for a fixed $\varepsilon>0$, 
we have that for every $t\ge \varepsilon$ and
$x\in X$,
\begin{align*}
\|\mathbb P_\alpha^\beta(t)x\|\le&  M\int_0^{\infty}
\frac{\tau}{t^{1-\frac{\alpha}{2}}} \Phi_{\frac{\alpha}{2}}(\tau)
e^{\omega \tau t^{\frac{\alpha}{2}}}\|x\| d\tau\le
M\|x\|\int_0^{\infty}
\tau \Phi_{\frac{\alpha}{2}}(\tau) e^{\omega \tau t^{\frac{\alpha}{2}}} d\tau\\
\le &M\|x\|\sum_{n=0}^\infty\frac{(\omega
t^{\frac{\alpha}{2}})^n}{n!} \int_0^{\infty}
\Phi_{\frac{\alpha}{2}}(\tau) \tau^{n+1}
d\tau \\
= & M\|x\|\sum_{n=0}^\infty\frac{(\omega
t^{\frac{\alpha}{2}})^n}{n!}
\frac{\Gamma(n+2)}{\Gamma(\frac{\alpha}{2}(n+1)+1)}\\
\le &M\|x\|\sum_{n=0}^\infty\frac{(\omega
t^{\frac{\alpha}{2}})^n}{\Gamma(\frac{\alpha}{2}(n+1))} =M\|x\|
E_{\frac{\alpha}{2},\frac{\alpha}{2}}(\omega
t^{\frac{\alpha}{2}})\le M_1 e^{t\omega^{\frac{2}{\alpha}}}\|x\|,
\end{align*}
for some constant $M_1\ge 0$, and this completes the proof.
\end{proof}

We notice that alternatively, one can also show that $\mathbb P_\alpha^\beta$ 
is exponentially bounded on $[0,\infty)$ by using the fact that 
$\mathbb S_\alpha^\beta$ is exponentially bounded and that 
$\mathbb P_\alpha^\beta(t)x=(g_{\alpha-1}\ast S_\alpha^\beta)(t)x$, $x\in X$, 
$t\ge 0$ (by Remark \ref{rem33}(a)).

If $B$ generates an exponentially bounded $\beta$-times integrated group 
$(U_\beta(t))$, then $A=B^2$ generates an exponentially bounded $\beta$-times 
integrated cosine family $(C_\beta(t))$ given by 
$C_\beta(t)=\frac{U_\beta(t)+U_\beta(-t)}{2}$. Moreover, operators that 
satisfy the estimate \eqref{e-chi} are generators of exponentially bounded 
integrated cosine families (see \cite[Theorem 2.2.4]{Ko2011} or \cite{Neu}). 
The corresponding situation for integrated semigroups is treated in 
\cite[Theorem 3.2.8]{ABHN01}.

Next, we show that we have a double subordination principle for
the families $\mathbb S_\alpha^\beta$ and $\mathbb P_\alpha^\beta$
in terms of the parameters $\alpha$ and $\beta$.

\begin{corollary}\label{cor-ISG}
Let $A$ be a closed linear operator on a Banach space $X$ and let
$1<\alpha\le 2,\beta\ge 0$. Then the following assertions hold.
\begin{itemize}
\item[(a)] If $A$ generates an $(\alpha,\alpha)^\beta$-resolvent family 
$\mathbb P_\alpha^\beta$, then it generates a $\frac{\beta}{2}$-times 
integrated semigroup $(T_{\frac{\beta}{2}}(t))$ such that
 $(g_1\ast T_{\frac{\beta}{2}})(t)$ is exponentially bounded and for 
every $x\in X$, and $t>0$,
\begin{equation*} 
T_{\frac{\beta}{2}}(t)x=\sigma t^{\sigma-1}\int_0^\infty
s\Phi_{\sigma}(s)\mathbb P_\alpha^\beta(s t^\sigma)x\,ds,\;\text{
where }\;\; \sigma:=\frac{1}{\alpha}.
\end{equation*}
\item[(b)] If $A$ generates an $(\alpha ,1)^\beta$-resolvent family 
$\mathbb S_\alpha^\beta$, then it generates a $\frac{\beta}{2}$-times 
integrated semigroup $(T_{\frac{\beta}{2}}(t))$ such that
 $(g_1\ast T_{\frac{\beta}{2}})(t)$ is exponentially bounded and for every $x\in X$, and $t\ge 0$,
\begin{equation*} 
T_{\frac{\beta}{2}}(t)x=\int_0^\infty \Phi_\sigma(s)\mathbb
S_\alpha^\beta(st^\sigma)x\,ds, \;\;\forall t\ge 0, \;\text{
where }\;\; \sigma:=\frac{1}{\alpha}.
\end{equation*}
\end{itemize}
\end{corollary}

The proof of Corollary \ref{cor-ISG} is a simple combination of
the proofs of Proposition \ref{prop-resol}, Theorem
\ref{th-resol-22} and Corollary \ref{cor-413}.

\begin{remark} \rm
(i) It follows from Theorem \ref{th-resol-22} and Corollary \ref{cor-ISG} that we have the following more general situation. Let $1<\alpha\le 2$ and $\beta\ge 0$ be given. If $A$ generates an $(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$, then $A$ also generates the  $(\alpha',1)^{\frac{\beta}{2}}$-resolvent family $S_{\alpha'}^{\frac{\beta}{2}}$ introduced in \cite{Ba01, KLW2013-2,KLW-1} for any $0<\alpha'\le 1$. More precisely, in \cite[Definition 4.2]{KLW-1}, for $0<\alpha'\le 1$ and $\beta\ge 0$, an $(\alpha',1)^{\beta}$-resolvent family associated to a closed linear operator $A$ defined on a Banach space $X$, has been defined to be a strongly continuous function $S_{\alpha'}^\beta:\; [0,\infty) \to\mathcal{L}(X)$ such that, $\|(g_1\ast S_{\alpha'}^\beta)(t)x\|\le Me^{\omega t}\|x||$, $x\in X$, $t\ge 0$, for some constants $M,\omega\ge 0$,   $\{\lambda^{\alpha'} \, : \,  \operatorname{Re}(\lambda)> \omega \} \subset \rho(A)$,   and
\begin{align*}
\lambda^{\alpha'-1}(\lambda^{\alpha'} -A)^{-1}x
=\lambda^{\alpha'\beta} \int_0^{\infty} e^{-\lambda t}
S_{\alpha'}^\beta(t)x dt,  \quad  \operatorname{Re}(\lambda) >
\omega, \quad x \in X.
\end{align*}
 In the same direction, we observe that a generator of an $(\alpha,1)$-resolvent
 family for $1<\alpha\le 2$ is already the generator of an analytic strongly
continuous semigroup.

(ii) We mention the following remarkable result obtained in \cite[Section 3]{Ba01}.
Let $A$ be a closed linear operator on a Banach space $X$. If $A$
generates a bounded analytic strongly continuous semigroup
$(T(t))$ of angle $\pi/2$, then $A$ generates an
$(\alpha,1)^0=(\alpha,1)$-resolvent family $\mathbb S_\alpha$ on
$X$ for every $1<\alpha<2$, and hence, generates an
$(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$ on
$X$ for every $1<\alpha<2$ and $\beta\ge 0$. But unfortunately,
there is no explicit representation of $\mathbb S_\alpha^\beta(t)$
in terms of $T(t)$.

(iii) In general, generators of resolvent families even in the case $\beta=0$
are not stable under bounded perturbations. In the case $\beta=0$, an
example in \cite[Example 2.24]{Ba01} shows that they need not be stable
by perturbations by multiple of the identity operator. Therefore the resolvent
families obtained through Corollary \ref{cor-413}  are of special interest
since they are stable under perturbations by multiple of the identities.
Other admissible perturbations have been studied, see e.g. \cite{ABHN01,Ko2011}
and the references therein.
\end{remark}

From Lemma \ref{lem-ex-b}, Theorem \ref{th-resol-22} and Corollary
\ref{cor-ISG} we derive the following result.

\begin{lemma}
Let $A$ be a closed linear operator on a Banach space $X$.
Let $1<\alpha\le 2$, $\beta\ge 0$ and $\mu>0$. Then the following
assertions hold.

(a) Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent family
$\mathbb P_{\alpha}^\beta$. Let $1\le \alpha'<\alpha$ and let
$\mathbb P_{\alpha' }^\beta$ be the $(\alpha',\alpha')^\beta$-resolvent
family generated by $A$, or the $\frac{\beta}{2}$-times integrated semigroup
$(T_{\frac{\beta}{2}}(t))$ generated by $A$. Then for every $x\in X$ and $t>0$,
\begin{gather}\label{E-CP}
\int_0^\infty\frac{\sigma
s}{t^{\sigma+1}}\Phi_\sigma(st^{-\sigma})(g_\mu\ast \mathbb
P_{\alpha }^\beta)(s)xds=(g_{\mu\sigma}\ast \mathbb
P_{\alpha' }^\beta)(t)x,\quad \sigma=\frac{\alpha'}{\alpha}, \\
\label{E-IS}
\int_0^\infty\frac{\sigma
s}{t^{\sigma+1}}\Phi_\sigma(st^{-\sigma})(g_\mu\ast \mathbb
P_{\alpha }^\beta)(s)xds=(g_{\mu\sigma}\ast T_{\frac{\beta}{2}
}(t)x,\quad  \sigma=\frac{1}{\alpha}.
\end{gather}

(b)  Assume that $A$ generates an $(\alpha,1)^\beta$-resolvent family
$\mathbb S_{\alpha}^\beta$. Let $1\le \alpha'<\alpha$, and let
$\mathbb S_{\alpha' }^\beta$ be the $(\alpha',1)^\beta$-resolvent
family generated by $A$, or the $\frac{\beta}{2}$-times integrated
semigroup $(T_{\frac{\beta}{2}}(t))$ generated by $A$.
Then for every $x\in X$ and $t>0$,
\begin{gather}\label{E-CS0}
\int_0^\infty\frac{1}{t^{\sigma}}\Phi_\alpha(st^{-\sigma})(g_\mu\ast
\mathbb S_{\alpha}^\beta)(s)xds=(g_{\mu\sigma}\ast \mathbb
S_{\alpha'}^\beta)(t)x,\quad \sigma:=\frac{\alpha'}{\alpha}, \\
\label{E-ISG0}
\int_0^\infty\frac{1}{t^{\sigma}}\Phi_\alpha(st^{-\sigma})(g_\mu\ast
\mathbb S_{\alpha}^\beta)(s)xds=(g_{\mu\sigma}\ast
T_{\frac{\beta}{2}}(t)x,\quad \sigma:=\frac{1}{\alpha}.
\end{gather}
\end{lemma}


\begin{proof}
Let $A$, $\alpha$, $\beta$ be as in the statement of the lemma and
let $x\in X$ and $\mu>0$.

(a) Assume that $A$ generates an $(\alpha,\alpha)^\beta$-resolvent
family $\mathbb P_{\alpha}^\beta$. Let $\omega$ be the real number
from the definition of $\mathbb P_\alpha^\beta$. Let $1\le
\alpha'<\alpha$. Using the Laplace transform, we have that for
$\operatorname{Re}(\lambda)>\omega$,
\begin{align}\label{J1}
 \widehat{(g_{\mu\sigma}\ast \mathbb
P_{\alpha'}^\beta)}(\lambda)x=\lambda^{-\mu\sigma}\lambda^{-\frac{\alpha'
\beta}{2}
}(\lambda^{\alpha'}-A)^{-1}x=\lambda^{-\mu\sigma-\frac{\alpha'\beta}{2}}(\lambda^{\alpha'}-A)^{-1}x.
\end{align}
On the other hand, using \eqref{eq2.3} and Fubini's theorem, we
obtain that for $\operatorname{Re}(\lambda)>\omega$,
\begin{equation} \label{J2}
\begin{aligned}
&\int_0^\infty e^{-\lambda t}\int_0^\infty\frac{\sigma
s}{t^{\sigma+1}}\Phi_\sigma(st^{-\sigma})(g_\mu\ast \mathbb
P_{\alpha }^\beta)(s)xds\,dt  \\
&=\int_0^\infty e^{-\lambda^{\sigma}s}(g_{\mu}\ast \mathbb P_\alpha^\beta)(s)xds \\
&=\lambda^{-\sigma(\mu+\frac{\alpha\beta}{2})}(\lambda^{\alpha\sigma}-A)^{-1}x \\
&=\lambda^{-\sigma\mu-\frac{\alpha^{\prime}\beta}{2}}(\lambda^{\alpha'}-A)^{-1}x.
\end{aligned}\end{equation}
In view of \eqref{J1} and \eqref{J2}, the equality \eqref{E-CP}
follows from the uniqueness theorem for the Laplace transform and
by continuity. The proof of \eqref{E-IS} follows the lines of the
proof of \eqref{E-CP}.

(b) Similarly, for $\operatorname{Re}(\lambda)>\omega$ (here $\omega$ be
the real number from the definition of $\mathbb S_\alpha^\beta$),
\begin{equation} \label{J3}
\begin{aligned}
 \widehat{(g_{\sigma\mu}\ast \mathbb
S_{\alpha'}^\beta)}(\lambda)x
&=\lambda^{-\sigma \mu}\lambda^{-\frac{\alpha' \beta}{2}
}\lambda^{\alpha'-1}(\lambda^{\alpha'}-A)^{-1}x \\
&=\lambda^{-\sigma\mu-\frac{\alpha'\beta}{2}}\lambda^{\alpha'-1}
(\lambda^{\alpha'}-A)^{-1}x.
\end{aligned}
\end{equation}

Using \eqref{eq2.3-2} and Fubini's theorem, we obtain for
$\operatorname{Re}(\lambda)>\omega$,
\begin{equation} \label{J4}
\begin{aligned}
&\int_0^\infty e^{-\lambda
t}\int_0^\infty\frac{1}{t^{\sigma}}\Phi_\alpha(st^{-\sigma})(g_\mu\ast
\mathbb S_{\alpha}^\beta)(s)xds\,dt\\
&= \lambda^{\sigma-1}\int_0^\infty e^{-\lambda^{\sigma}t}
(g_{\alpha\mu}\ast \mathbb S_\alpha^\beta)(s)xds \\
&= \lambda^{\sigma-1}\lambda^{-\mu\sigma-\sigma\frac{\alpha\beta}{2})}
\lambda^{\sigma(\alpha-1)}(\lambda^{\sigma\alpha}-A)^{-1}x \\
&= \lambda^{\sigma-1}\lambda^{-\mu\sigma-\frac{\alpha'\beta}{2})}
 \lambda^{\alpha'-\sigma}(\lambda^{\alpha'}-A)^{-1}x \\
&= \lambda^{-\sigma\mu-\frac{\alpha'\beta}{2}}\lambda^{\alpha'-1}
(\lambda^{\alpha'}-A)^{-1}x.
\end{aligned}
\end{equation}
Using \eqref{J3} and \eqref{J4}, the equality \eqref{E-CS0} also
follows from the uniqueness theorem for the Laplace transform and
by continuity. The proof of \eqref{E-ISG0} also follows the lines
of the proof of \eqref{E-CS0}.
\end{proof}


The following result on the regularity properties of the family
$\mathbb S_\alpha^\beta$ is crucial and will be used several times
in the subsequent sections to obtain our main results.

\begin{lemma}\label{lem-reg}
 Let $A$ be a closed linear operator with domain $D(A)$ defined
on a Banach space $X$. Let $1< \alpha\le 2, \beta\ge 0$,
$k:=\lceil \frac{\alpha\beta}{2}\rceil$, $n:=\lceil \beta\rceil$
and assume that $A$ generates an $(\alpha,1)^\beta$-resolvent
family $\mathbb S_\alpha^\beta$.  Then the following properties
hold.

(a) Let $m\in\mathbb{N}\cup\{0\}$. Then for every $x\in D(A^{m+1})$ and $t\ge 0$,
\begin{align}\label{Eq-Tay}
\mathbb
S_\alpha^\beta(t)x=\sum_{j=0}^{m}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
+\int_0^tg_{\alpha
(m+1)}(t-s)\mathbb S_\alpha^\beta(s)A^{m+1}x\,ds.
\end{align}

(b) For every $x\in D(A^{n+1})$, the  map
$t\mapsto (g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x$
belongs to the space $C^k([0,\infty);D(A))\cap C^{k+1}([0,\infty);X)$ and 
for every $t\ge 0$,
\begin{gather}\label{eqD}
\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)x\right]=\sum_{j=0}^{n-1}g_{\alpha
j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb
S_\alpha^\beta)(t)A^nx, \\
\label{eq-der3}
\frac{d^{k+1}}{dt^{k+1}}\left[(g_{k-\frac{\alpha\beta}{2}}\ast
\mathbb S_\alpha^\beta)(t)x\right]
=\sum_{j=1}^{n}g_{\alpha j}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})+\alpha-1}
 \ast \mathbb S_\alpha^\beta)(t)A^{n+1}x.
\end{gather}
In particular,
\begin{gather}\label{E111}
\frac{d^{j}}{dt^{j}}\left[g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta\right](0)x=0,\;j=0,1,\ldots, k-1,\;\;\;
\frac{d^{k}}{dt^{k}}\left[g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta\right](0)x=x, \\
\label{eq-der0}
\frac{d^{k+1}}{dt^{k+1}}\left[g_{k-\frac{\alpha\beta}{2}}\ast\mathbb
S_\alpha^\beta\right](0)x=0,\quad \frac{d^{k+1}}{dt^{k+1}}\left[g_1\ast
g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta\right](0)x=x.
\end{gather}

(c) In general, for every $x\in D(A^{n+1-i})$, $i=0,1,\ldots,n$,
the  mapping $t\mapsto (g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha i}
\ast \mathbb S_\alpha^\beta)(t)x$ belongs to $C^k([0,\infty);D(A))$
and for every $t\ge 0$,
\begin{equation} \label{eqDD}
\begin{aligned}
&\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}\ast
g_{\alpha i}\ast \mathbb
S_\alpha^\beta)(t)x\right]\\
&=\sum_{j=0}^{n-i}g_{\alpha j+1+\alpha
i}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast\mathbb
S_\alpha^\beta)(t)A^{n+1-i}x.
\end{aligned}
\end{equation}

(d) For every $x\in D(A^{n})$, the  mapping
$t\mapsto (g_{k-\frac{\alpha\beta}{2}}\ast\mathbb S_\alpha^\beta)(t)x$ belongs 
to the class  $C^k([0,\infty);X)$ and the equalities \eqref{eqD} 
and \eqref{E111} hold.

(e) In general, for every $x\in D(A^{n-i})$, $i=0,1,\ldots, n$,
the  mapping $t\mapsto (g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha i}
\ast\mathbb S_\alpha^\beta)(t)x$ belongs to $C^k([0,\infty);X)$ and for every
$t\ge 0$,
\begin{equation} \label{eqDD-22}
\begin{aligned}
&\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}\ast
g_{\alpha i}\ast \mathbb
S_\alpha^\beta)(t)x\right]\\
&=\sum_{j=0}^{n-i}g_{\alpha j+1+\alpha
i}(t)A^jx+A(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast\mathbb
S_\alpha^\beta)(t)A^{n-i}x.
\end{aligned}
\end{equation}
\end{lemma}


\begin{proof}
 Let $A$ be a closed linear operator with domain $D(A)$ defined
on a Banach space $X$. Let $1< \alpha\le 2, \beta\ge 0$  and set
$k:=\lceil \frac{\alpha\beta}{2}\rceil$, $n:=\lceil \beta\rceil$.
Note that $k\le n$. Assume that $A$ generates an
$(\alpha,1)^\beta$-resolvent family $\mathbb S_\alpha^\beta$.

(a) We prove \eqref{Eq-Tay} by induction. If $m=0$, then for every
$x\in D(A)$, the equality \eqref{Eq-Tay} reads
\[
\mathbb S_\alpha^\beta(t)x=g_{\frac{\alpha\beta}{2}+1}(t)x
+\int_0^tg_{\alpha }(t-s)\mathbb S_\alpha^\beta(s)Ax\,ds,\quad \forall t\ge 0
\]
which is given by Lemma \ref{le2.3-2}(b). Assume that
\eqref{Eq-Tay} holds for $m-1$ for some $m\in\mathbb{N}$. Now, let $x\in
D(A^{m+1})\subset D(A^m)$. Then using Lemma \ref{le2.3-2}(b), we
have that for every $t\ge 0$,
\begin{align*}
\mathbb S_\alpha^\beta(t)x
=&\sum_{j=0}^{m-1}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
 +(g_{\alpha m}\ast\mathbb S_\alpha^\beta)(t)A^{m}x\\
=&\sum_{j=0}^{m-1}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
 +A^m(g_{\alpha m}\ast\mathbb S_\alpha^\beta)(t)x\\
=&\sum_{j=0}^{m-1}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
 +A^mg_{\alpha m}\ast\Big(g_{\frac{\alpha\beta}{2}+1}x
 +g_\alpha\ast \mathbb S_\alpha^\beta Ax\Big)(t)\\
=&\sum_{j=0}^{m-1}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
 +g_{\alpha(\frac{\beta}{2}+m)+1}(t)A^mx+(g_{\alpha(m+1)}
 \ast \mathbb S_{\alpha\beta})(t)A^{m+1}x\\
=&\sum_{j=0}^{m}g_{\alpha(\frac{\beta}{2}+j)+1}(t)A^jx
 +(g_{\alpha(m+1)}\ast\mathbb S_\alpha^\beta)(t)A^{m+1}x.
\end{align*}
We conclude that the equality \eqref{Eq-Tay} holds and this
completes the proof of part (a).

(b) Let $x\in D(A^{n+1})$. Then using \eqref{Eq-Tay} with $m=n$ we
get that for every $t\ge 0$,
\[
(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x
=\sum_{j=0}^{n}g_{k+\alpha
j+1}(t)A^jx+(g_{\alpha(n+1)+k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)A^{n+1}x.
\]
Therefore, using Lemma \ref{le2.3-2}(b), we have that for all
$t\ge 0$,
\begin{align*}
\frac{d^k}{dt^k}
\big[(g_{k-\frac{\alpha\beta}{2}}\ast\mathbb S_\alpha^\beta)(t)x\big]
=&\sum_{j=0}^{n}g_{\alpha j+1}(t)A^jx+(g_{\alpha( n+1)
 -\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)A^{n+1}x\\
=&\sum_{j=0}^{n}g_{\alpha j+1}(t)A^jx
 +\left(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast \mathbb S_\alpha^\beta\right)
 (t)A^{n+1}x\\
=&\sum_{j=0}^{n}g_{\alpha j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast
 (\mathbb S_\alpha^\beta-g_{\frac{\alpha\beta}{2}+1}))(t)A^nx\\
=&\sum_{j=0}^{n}g_{\alpha j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast
 \mathbb S_\alpha^\beta)(t)A^nx-g_{\alpha n+1}(t)A^nx\\
=&\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb
S_\alpha^\beta)(t)A^nx,
\end{align*}
and we have shown \eqref{eqD}. Since $A^nx\in D(A)$, it follows
from \eqref{eqD} and Lemma \ref{le2.3-2} that
$\frac{d^k}{dt^k}[(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)x]\in C([0,\infty);D(A))$. Hence,
$ (g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)x\in C^k([0,\infty);D(A))$. Since $g_1(0^+)=1$ and
$g_{\alpha j+1}(0^+)=0$  for every $j=1,2,\ldots, n-1$, the
equalities in \eqref{E111} follow from \eqref{eqD}.

By Remark \ref{rem33} and Lemma \ref{lem-deri}, $A$ generates an
$(\alpha,\alpha)^\beta$-resolvent family $\mathbb P_\alpha^\beta$
and for every $x\in D(A)$, $ \mathbb S_\alpha^\beta(t)x\in
C([0,\infty);D(A))\cap C^1((0,\infty);X)$. Now, let $x\in
D(A^{n+1})$. We have to show that
$(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x\in
C^{k+1}([0,\infty);X)$ and \eqref{eq-der0} holds. It follows from
\eqref{eqD}, \eqref{eq-der} and the fact that $\mathbb
S_\alpha^\beta(t)A^nx\in C([0,\infty);D(A))\cap
C^1((0,\infty);X)$, that for every $t\ge 0$,
\begin{align*}
&\frac{d^{k+1}}{dt^{k+1}}(g_{k-\frac{\alpha\beta}{2}}\ast
\mathbb S_\alpha^\beta)(t)x\\
&=\sum_{j=1}^{n-1}g_{\alpha j}(t)A^jx
 +\big[(g_{\alpha(n-\frac{\beta}{2})}\ast (\mathbb S_\alpha^\beta)')(t)A^nx\big] \\
&=\sum_{j=1}^{n-1}g_{\alpha j}(t)A^jx+\left[g_{\alpha(n-\frac{\beta}{2})}\ast
 \left(g_{\frac{\alpha\beta}{2}} A^nx+\mathbb P_\alpha^\beta A^{n+1}x\right)(t)\right] \\
&=\sum_{j=1}^{n}g_{\alpha j}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb P_\alpha^\beta)(t)A^{n+1}x \\
&=\sum_{j=1}^{n}g_{\alpha
j}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})+\alpha-1}\ast \mathbb
S_\alpha^\beta)(t)A^{n+1}x\in C([0,\infty);X),
\end{align*}
and we have shown \eqref{eq-der3}. Therefore,
$(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x\in
C^{k+1}([0,\infty);X)$. It also follows from \eqref{eq-der3} that
\[\frac{d^{k+1}}{dt^{k+1}}\left[g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta\right](0)x=0.\]
Now, using \eqref{E111} we get that
\[
\frac{d^{k+1}}{dt^{k+1}}\left[g_1\ast g_{k-\frac{\alpha\beta}{2}}
\ast \mathbb S_\alpha^\beta\right](0)x
=\frac{d^{k}}{dt^{k}}\left[g_{k-\frac{\alpha\beta}{2}}\ast
\mathbb S_\alpha^\beta\right](0)x=x\,.
\]
This completes the proof of part (b).

(c) Let $x\in D(A^{n+1-i})$, $i=0,1,\ldots,n$. Proceeding as in
the proof of part (b) we get that for every $t\ge 0$,
\begin{align*}
(g_{\alpha i}\ast g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)x=\sum_{j=0}^{n-i}g_{k+\alpha j+\alpha
i+1}(t)A^jx+(g_{k+\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast
\mathbb S_\alpha^\beta)(t)A^{n+1-i}x.
\end{align*}
This implies that for every $t\ge 0$,
\begin{align*}
&\frac{d^k}{dt^k}\left[(g_{\alpha i}\ast
g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta)(t)x\right]\\
&=\sum_{j=0}^{n-i}g_{\alpha j+\alpha
i+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast\mathbb
S_\alpha^\beta)(t)A^{n+1-i}x.
\end{align*}
Using Lemma \ref{le2.3-2}, the preceding equality shows that
$ (g_{\alpha i}\ast g_{k-\frac{\alpha\beta}{2}}\ast
\mathbb S_\alpha^\beta)(t)x\in C^k([0,\infty);D(A))$ and one has
the equality \eqref{eqDD}.

(d) Let $x\in D(A^n)$. Proceeding as in part (b), we also get the
equality \eqref{eqD} and this implies that
$(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x\in
C^k([0,\infty);X)$ and \eqref{E111} holds.

(e) Let $x\in D(A^{n-i}), i=0,1,\ldots, n$. Proceeding as in part
(c), we get that $(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha
i}\ast \mathbb S_\alpha^\beta)(t)x\in C^k([0,\infty);X)$ and the
equality \eqref{eqDD} holds. The proof of the lemma is complete.
\end{proof}



\section{Resolvent families and the regularized abstract Cauchy problem}\label{sec4}

In this section we show that the above defined resolvent family
$\mathbb S_\alpha^\beta$ is necessary and sufficient to solve the
regularized abstract Cauchy problem
\begin{equation}\label{E1}
\begin{gathered}
\mathbb D_t^\alpha v(t)=Av(t)+g_{\frac{\alpha\beta}{2}+1}(t)x,\quad
t> 0,\;1<\alpha\le 2,\;\beta\ge 0,\\
v(0)=v'(0)=0,
\end{gathered}
\end{equation}
where $A$ is a closed linear operator with domain $D(A)$ defined
on a Banach space $X$. By a classical solution of \eqref{E1} we
mean a function $v\in C([0,\infty);D(A))\cap C^{1}([0,\infty);X)$
such that  $(g_{2-\alpha}\ast v)\in C^2([0,\infty);X)$ and
\eqref{E1} is satisfied.

The following is the  main result of this section.

\begin{theorem}\label{th-51}
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$. Let $1<\alpha\le 2$ and $\beta\ge 0$. Then the
following assertions are equivalent.
\begin{itemize}
\item[(i)] The operator $A$ generates an $(\alpha,1)^\beta$-resolvent family
 $\mathbb S_{\alpha}^\beta$ on $X$.

\item[(ii)] For all $x\in X$, there exists a unique classical solution $v$
 of Problem \eqref{E1} such that $(g_{2-\alpha}\ast v')(t)$ is exponentially bounded.
\end{itemize}
\end{theorem}

\begin{proof}
Let $A$, $\alpha$ and $\beta$  be as the statement of the theorem.

(i) $\Rightarrow$ (ii): Assume that $A$ generates an
$(\alpha,1)^\beta$-resolvent family $\mathbb S_{\alpha}^\beta$ on
$X$ and let $x\in X$. Define
\[
v(t):=(g_\alpha\ast \mathbb S_\alpha^\beta)(t)x
=\int_0^tg_{\alpha}(t-s)\mathbb S_\alpha^\beta(s)x\,ds,\quad t\ge 0.
\]
Then $v(0)=0$ and by Lemma \ref{le2.3-2} we have that
$v\in C([0,\infty);D(A))$. Since $v'(t)=(g_{\alpha-1}\ast \mathbb
S_\alpha^\beta)(t)x$, we have that $v \in C^{1}([0,\infty);X)$ and
$v'(0)=0$. Since for every $t\ge 0$,
\[
(g_{2-\alpha}\ast v)(t)=(g_{2-\alpha}\ast g_\alpha\ast
\mathbb S_\alpha^\beta)(t)x=(g_2\ast\mathbb  S_\alpha^\beta)(t)x,
\]
it follows that $(g_{2-\alpha}\ast v)\in C^{2}([0,\infty);X)$.
Since $v(0)=v'(0)=0$, it follows from \eqref{RLC} and
\eqref{conv1} that for every $t\ge 0$,
\begin{align*}
\mathbb D_t^\alpha v(t)
=&(g_{2-\alpha}\ast v'')(t)
 =\frac{d^2}{dt^2}\left[(g_{2-\alpha}\ast v)(t)\right]\\
=&\frac{d^2}{dt^2}\left[(g_{2}\ast \mathbb S_\alpha^\beta)(t)x\right]
 =\mathbb S_\alpha^\beta(t)x\\
=&A(g_{\alpha}\ast\mathbb S_\alpha^\beta)(t)x +g_{\frac{\alpha\beta}{2}+1}(t)x
=Av(t)+g_{\frac{\alpha\beta}{2}+1}(t)x.
\end{align*}
Hence, $v$ is a classical solution of \eqref{E1}. Since $(g_1\ast
\mathbb S_\alpha^\beta)(t)$ is exponentially bounded and for every
$x\in X$, $t\ge 0$,
\[
(g_{2-\alpha}\ast v')(t)
=(g_{2-\alpha}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta)(t)x
=(g_1\ast \mathbb S_\alpha^\beta)(t)x,
\]
 it follows that $(g_{2-\alpha}\ast v')(t)$ is exponentially bounded.
Assume that \eqref{E1} has two classical solutions $v_1$ and $v_2$
and set $V:=v_1-v_2$. Then $V\in C([0,\infty);D(A))\cap
C^{1}([0,\infty);X)$, $V(0)=V'(0)=0$, $(g_{2-\alpha}\ast
V)\in C^2([0,\infty);X)$, $(g_{2-\alpha}\ast V)(t)$ is
exponentially bounded and $\mathbb D_t^\alpha V(t)=AV(t)$ for
every $t> 0$. Taking the Laplace transform, we get that for
$\operatorname{Re}(\lambda)>\omega$ (where $\omega$ is the real number
from the above mentioned exponential boundedness),
$(\lambda^\alpha-A)\widehat{V}(\lambda)=0$. Since
$(\lambda^\alpha-A)$ is invertible, we have that
$\widehat{V}(\lambda)=0$. By the uniqueness theorem for the
Laplace transform and by continuity, we get that $V(t)=0$ for
every $t\ge 0$. We have shown uniqueness of solutions and this
completes the proof of part (ii).

(ii) $\Rightarrow$ (i): For $x\in X$, we let
$S_{\alpha,\beta}(t)x:=\mathbb D_t^\alpha v(t,x)$  where $v(t,x)$
is the  unique classical solution of \eqref{E1}.  Using
\eqref{RLC} and the fact that $v(0)=0=v'(0)$ we get that for
every $t\ge 0$,
\begin{align*}
(g_\alpha\ast S_{\alpha,\beta})(t)x=(g_{\alpha}\ast \mathbb
D_t^\alpha v)(t)=v(t,x)-v(0,x)-v'(0,x)t=v(t,x).
\end{align*}
Hence, $(g_\alpha\ast S_{\alpha,\beta})(t)x\in D(A)$ for every
$x\in X$, $t\ge 0$, and one has the equality
\begin{equation}\label{e222}
A(g_\alpha\ast
S_{\alpha,\beta})(t)x+g_{\frac{\alpha\beta}{2}+1}(t)x=Av(t,x)+g_{\frac{\alpha\beta}{2}+1}(t)x=S_{\alpha,\beta}(t)x.
\end{equation}
By the closed graph theorem we also have that
$S_{\alpha,\beta}(t)\in\mathcal L(X)$ for $t\ge 0$ and we note
that $S_{\alpha,\beta}(t)$ is strongly continuous on $[0,\infty)$.
Since by hypothesis $(g_{2-\alpha}\ast v')(t)$ is
exponentially bounded and given that for every $x\in X$, $t\ge 0$,
\[(g_1\ast S_{\alpha,\beta})(t)x=(g_1\ast g_{2-\alpha}\ast v'')(t)=(g_{2-\alpha}\ast v^{\prime})(t),\]
we have that $(g_1\ast S_{\alpha,\beta})(t)x$ is exponentially
bounded. By the uniform exponential boundedness principle
\cite[Lemma 3.2.14]{ABHN01}, we have that there exist some
constants $M,\omega\ge 0$ such that
\begin{equation}\label{vv}
\|(g_{2-\alpha}\ast v')(t)\|=\|(g_1\ast
S_{\alpha,\beta})(t)x\|\le Me^{\omega t}, \quad t\ge 0, \;x\in X.
\end{equation}
Taking the Laplace transform on both sides of the equality
\eqref{e222} we get that for $\operatorname{Re}(\lambda)>\omega$ (where
$\omega$ is the real number from the above mentioned exponential
boundedness),
\begin{align*}
A\lambda^{-\alpha}\widehat{S}_{\alpha,\beta}(\lambda)x
-\widehat{S}_{\alpha,\beta}(\lambda)x=-\lambda^{-\frac{\alpha\beta}{2}-1}x.
\end{align*}
Multiplying the preceding equality by $\lambda^\alpha$ we get that
\begin{align*}
(\lambda^\alpha-A)\widehat{S}_{\alpha,\beta}(\lambda)x
=\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}x.
\end{align*}
The above equality shows that $(\lambda^\alpha-A)$ is
surjective. To prove injectivity, suppose that
$(\lambda^\alpha-A)x=0$ for some $x\in D(A)$ and
$\operatorname{Re}(\lambda)>\omega$, that is, $Ax=\lambda^\alpha x$ for
$\operatorname{Re}(\lambda)>\omega$. It is enough to consider that
$Ax=\lambda^\alpha x$ for $\lambda$ real and $\lambda>\omega$.
Then setting $v(t)=(g_{\frac{\alpha\beta}{2} +\alpha}\ast
\widetilde E)(t)x$  where $\widetilde
E(t)x=E_{\alpha,1}(\lambda^\alpha t^\alpha)x$, we prove that $v$
is a solution of Equation \eqref{E1}. Obviously $v\in
C([0,\infty);D(A))\cap C^1([0,\infty);X)$ and $(g_{2-\alpha}\ast
v)\in C^2([0,\infty);X)$. Using \eqref{new}, we have that for
every $t>0$,
\begin{align*}
\mathbb D_t^\alpha v(t)
=&g_{2-\alpha}\ast\frac{d^2}{dt^2}\left[(g_{\frac{\alpha\beta}{2}
 +\alpha}\ast \widetilde E)(t)x\right] \\
=&(g_{\frac{\alpha\beta}{2}}\ast \widetilde E)(t)x
 =g_{\frac{\alpha\beta}{2}}\ast (g_1+\lambda^\alpha g_\alpha\ast \widetilde E))(t)x\\
=& g_{\frac{\alpha\beta}{2}+1}(t)x+(g_{\frac{\alpha\beta}{2}+\alpha}\ast \widetilde E)
 (t)\lambda^\alpha x
 = g_{\frac{\alpha\beta}{2}+1}(t)x+(g_{\frac{\alpha\beta}{2}+\alpha}\ast \tilde E)(t)
 A x\\
=&g_{\frac{\alpha\beta}{2}+1}(t)x+A(g_{\frac{\alpha\beta}{2}+\alpha}\ast
\widetilde E)(t)x= g_{\frac{\alpha\beta}{2}+1}(t)x+Av(t).
\end{align*}
We have shown that $v$ is a solution of Equation \eqref{E1}. Since
all the solutions $v$ of  Equation \eqref{E1} satisfy the estimate
\eqref{vv}, we must have this estimate for the solution
$v(t)=(g_{\frac{\alpha\beta}{2} +\alpha}\ast \widetilde E)(t)x$
just found. But using \eqref{mm1} we have that
\[
\widetilde E(t)=\sum_{n=0}^\infty\frac{\lambda^{\alpha n}
t^{\alpha n}}{\Gamma(\alpha n+1)}
\]
which gives
\begin{align*}
(g_{2-\alpha}\ast v')(t)
&=(g_{\frac{\alpha\beta}{2}+1}\ast
\tilde E)(t)x=t^{\frac{\alpha\beta}{2}+1}\sum_{n=0}^\infty
\frac{\lambda^{\alpha n}t^{\alpha n}}{\Gamma(\alpha
n+\frac{\alpha\beta}{2}+2)} \\
&= t^{\frac{\alpha\beta}{2}+1}E_{\alpha,\frac{\alpha\beta}{2}+2}(\lambda^\alpha
t^\alpha)x,
\end{align*}
and hence by  \eqref{Mit-E}, $\|(g_{2-\alpha}\ast
v')(t)\|\le Me^{\lambda t}\|x\|$ and this estimate is sharp.
Therefore we can only have the estimate \eqref{vv} if $x=0$. We
have shown that $(\lambda^\alpha-A)$ is injective, hence is
invertible and
\begin{align*}
\widehat{S}_{\alpha,\beta}(\lambda)x=\lambda^{-\frac{\alpha\beta}{2}}\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}x,
\end{align*}
that is, for $\operatorname{Re}(\lambda)>\omega$, and $x\in X$,
\begin{align*}
\lambda^{\alpha-1}(\lambda^\alpha-A)^{-1}x=\lambda^{\frac{\alpha\beta}{2}}\int_0^\infty
e^{-\lambda t}  S_{\alpha,\beta}(t)x\,dt.
\end{align*}
Hence, $A$ generates an $(\alpha,1)^\beta$-resolvent family
$\mathbb S_\alpha^\beta$ and by the uniqueness theorem for the
Laplace transform and by continuity we have that $\mathbb
S_\alpha^\beta(t)x=S_{\alpha,\beta}(t)x$ for every $x\in X$, $t\ge
0$. We have shown the assertion (i) and the proof is finished.
\end{proof}

\begin{remark}\rm
(a) We note that in Theorem \ref{th-51}, the assertion that
$(g_{2-\alpha}\ast v')(t)$ is exponentially bounded agrees with the
limiting cases $\alpha=1$ in which the conclusion reads $(g_1\ast v')(t)=v(t)$
is exponentially bounded (see e.g. \cite[Theorem 3.2.13]{ABHN01}), and $\alpha=2$,
in which we have that $v'(t)$ is exponentially bounded. An example showing that
the exponential boundedness assumption cannot be omitted is included
in \cite[Remark 3.2.15(b)]{ABHN01} for the limiting case $\alpha=1$.

(b) We mention that if the family $\mathbb S_\alpha^\beta$ is exponentially bounded,
then the solution $v$ in Theorem \ref{th-51} is exponentially bounded as well.
\end{remark}

\section{Resolvent families and the homogeneous abstract Cauchy problem}\label{sec-ho-p}

In this section we use the above defined resolvent families to
investigate the existence, regularity and the representation of
solutions of  the homogeneous abstract Cauchy problem
\begin{equation}\label{ACP-H2}
\begin{gathered}
\mathbb D_t^\alpha u(t)=Au(t),\quad t>0,\;1<\alpha\le 2,\\
u(0)=x,\quad  u'(0)=y,
\end{gathered}
\end{equation}
where $A$ is a closed linear operator with domain $D(A)$ defined
on a Banach space $X$ and $x,y$ are given vectors in $X$.

\begin{definition} \rm
A function $u\in C([0,\infty);D(A))\cap C^1([0,\infty);X)$ is said
to be a classical solution of Problem \eqref{ACP-H2} if
$g_{2-\alpha}\ast\left(u-u(0)-u'(0)g_2\right)\in
C^2([0,\infty);X)$  and \eqref{ACP-H2} is satisfied.
\end{definition}

We  adopt the following definition of mild solutions.

\begin{definition} \rm
A function $u\in C([0,\infty);X)$ is said to be a mild solution of
\eqref{ACP-H2} if $I_t^\alpha u(t):=(g_\alpha\ast u)(t)\in D(A)$
for every $t\ge 0$,  and
\begin{align*}
u(t)=x+ty+ A\int_0^tg_\alpha(t-s)u(s)\,ds,\;\;\forall  t\ge 0.
\end{align*}
\end{definition}

We have the following uniqueness result.

\begin{proposition}\label{puniq-2-2}
Let $A$ be a closed and linear operator with domain $D(A)$ defined
on a Banach space $X$ and let $1<\alpha\le 2$. Then the following
assertions hold.
\begin{itemize}
\item[(a)] If $u$ is a classical solution of \eqref{ACP-H2}, then
it is a mild solution of \eqref{ACP-H2}.

\item[(b)] If $(\lambda^\alpha-A)$ is invertible for $\operatorname{Re}(\lambda)$
 large enough, and if a mild solution $u$ exists and $(g_1\ast u)(t)$ is
 exponentially bounded, then it is unique.
\end{itemize}
\end{proposition}

\begin{proof}
Let $1<\alpha\le 2$ and let $A$ be a closed linear operator with
domain $D(A)$ defined on a Banach space $X$.

(a) Let $u$ be a classical solution of \eqref{ACP-H2}. Since $u\in
C([0,\infty);D(A))$, we have that $(g_\alpha\ast u)(t)\in
C([0,\infty);D(A))$. Since $\mathbb D_t^\alpha u(t)=Au(t)$, that
is, $(g_{2-\alpha}\ast u'')(t)=Au(t)$, we have that
$(g_\alpha\ast g_{2-\alpha}\ast
u'')(t)=A(g_\alpha\ast u)(t)$, i.e., $(g_{2}\ast
u'')(t)=A(g_\alpha\ast u)(t)$. Hence,
$u(t)-u(0)-tu'(0)=A(g_\alpha\ast u)(t)$ for every $t\ge 0$
and we have shown that $u$ is a mild solution of \eqref{ACP-H2}.

(b) Assume that \eqref{ACP-H2} has two mild solutions $u$ and $v$
and set $U:=u-v$. Then $U\in C([0,\infty);X)$, $(g_\alpha\ast
U)(t)\in D(A)$ for every $t\ge 0$ and $U(t)=A(g_\alpha\ast U)(t)$.
Taking the Laplace transform, we get that
$(I-\lambda^{-\alpha}A)\widehat{U}(\lambda)=0$ for
$\operatorname{Re}(\lambda)>\omega$ (where $\omega\ge 0$ is the real
number from the exponential boundedness of $(g_1\ast u)(t)$).
Since $(I-\lambda^{-\alpha}A)$ is invertible, we have that
$\widehat{U}(\lambda)=0$. By the uniqueness theorem for the
Laplace transform and by continuity, we get that $U(t)=0$ for
every $t\ge 0$. Hence, $u(t)=v(t)$ for every $t\ge 0$. The proof
is finished.
\end{proof}

\begin{remark} \rm
We mention that to prove the existence of solutions of
Problem \eqref{ACP-H2}, we proceed by direct construction and make
minimal use of the Laplace transform.
\end{remark}

The following result is the main result of this section.

\begin{theorem}\label{the-m2}
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$. Let $ 1<\alpha\le 2, \beta\ge 0$ and set
$n:=\lceil \beta\rceil$, $k:=\lceil \frac{\alpha\beta}{2}\rceil$.
Assume that $A$ generates an $(\alpha,1)^{\beta}$-resolvent family
$\mathbb S_\alpha^{\beta}$. Then the following assertions hold.
\begin{itemize}

\item[(a)] For  every $x,y\in D(A^{n+1})$, the function
$u(t):=D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t) x
+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t) y$
is the unique classical solution of \eqref{ACP-H2}.

\item[(b)] For  every $x,y\in D(A^n)$, the function
$u(t):=D_t^{\frac{\alpha\beta}{2}} \mathbb S_\alpha^\beta(t) x
+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta) (t)y$
is the unique mild solution of \eqref{ACP-H2}.
\end{itemize}
\end{theorem}

\begin{proof}
Let $A$, $\alpha$, $\beta$, $n:=\lceil \beta\rceil$ and $k:=\lceil
\frac{\alpha\beta}{2}\rceil$ be as in the statement of the
theorem.  First we prove existence of classical and mild
solutions.

(a) Let $x,y\in D(A^{n+1})$ and set
$u(t):=D_t^{\frac{\alpha\beta}{2}} \mathbb S_\alpha^\beta(t)
x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)
(t)y$. It follows from Lemma \ref{lem-reg} that $u\in
C([0,\infty);D(A))\cap C^1([0,\infty);X)$, $u(0)=x$ and $u'(0)=y$.
Since $u(0)=x$, $u'(0)=y$, using Lemma \ref{lem-reg} and Lemma
\ref{le2.3-2},  we have that for every $t\ge 0$,
\begin{equation} \label{D1}
\begin{aligned}
&g_{2-\alpha}\ast (u-u(0)-u'(0)g_2)(t)\\
&=g_{2-\alpha}\ast\Big[ \sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^jx
 +(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb S_\alpha^\beta)(t)A^nx-x\Big] \\
&\quad +g_{2-\alpha}\ast\Big[\sum_{j=0}^{n-1}g_{\alpha j+2}(t)A^jy
 +(g_{\alpha(n-\frac{\beta}{2})+1}\ast \mathbb S_\alpha^\beta)(t)A^ny-ty\Big] \\
&= \sum_{j=1}^{n-1}g_{\alpha j+3-\alpha}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})
 +2-\alpha}\ast \mathbb S_\alpha^\beta)(t)A^nx \\
&\quad +\sum_{j=1}^{n-1}g_{\alpha j+4-\alpha}(t)A^jy
 +(g_{\alpha(n-\frac{\beta}{2})+3-\alpha}\ast\mathbb S_\alpha^\beta)(t)A^ny \\
&=\sum_{j=1}^{n}g_{\alpha j+3-\alpha}(t)A^jx
 +(g_{\alpha(n-\frac{\beta}{2})+2}\ast \mathbb S_\alpha^\beta)(t)A^{n+1}x \\
&\quad +\sum_{j=1}^{n}g_{\alpha j+4-\alpha}(t)A^jy
 +(g_{\alpha(n-\frac{\beta}{2})+3}\ast \mathbb S_\alpha^\beta)(t)A^{n+1}y.
\end{aligned}
\end{equation}
Using \eqref{D1} and Lemma \ref{lem-reg} we get that for every
$t\ge 0$,
\begin{align*}
&\frac{d^2}{dt^2}\Big[g_{2-\alpha}\ast (u-u(0)-u'(0)g_2)\Big](t) \\
&= \sum_{j=1}^{n}g_{\alpha j+1-\alpha}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}
 \ast \mathbb S_\alpha^\beta)(t)A^{n+1}x\\
&+\sum_{j=1}^{n}g_{\alpha j+2-\alpha}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}
\ast \mathbb S_\alpha^\beta)(t)A^{n+1}y
\in C([0,\infty);X).
\end{align*}
Hence, $g_{2-\alpha}\ast (u-u(0)-u'(0)g_2)\in C^2([0,\infty); X)$.
We have to show that  $u$ satisfies \eqref{ACP-H2}. Using
\eqref{eq-der3} in Lemma \ref{lem-reg}, we get that for every
$t\ge 0$,
\begin{align*}
\mathbb D_t^\alpha u(t)
=&\mathbb D_t^\alpha
D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t)x+
\mathbb D_t^\alpha D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t)y\\
=&g_{2-\alpha}\ast\Big[\frac{d^{k+2}}{dt^{k+2}}\left(g_{k-\frac{\alpha\beta}{2}}\ast
\mathbb S_\alpha^\beta\right)(t)x+
\frac{d^{k+2}}{dt^{k+2}}\left(g_{k-\frac{\alpha\beta}{2}}\ast g_1\ast
 \mathbb S_\alpha^\beta\right)(t)y\Big]\\
=&g_{2-\alpha}\ast\frac{d}{dt}\Big[\sum_{j=1}^ng_{\alpha j}(t)A^jx
 +(g_{\alpha(n-\frac{\beta}{2})+\alpha-1}\ast \mathbb S_\alpha^\beta)
 (t)A^{n+1}x\Big]\\
&+g_{2-\alpha}\ast\Big[\sum_{j=1}^ng_{\alpha j}(t)A^jy
 +(g_{\alpha(n-\frac{\beta}{2})+\alpha-1}\ast \mathbb S_\alpha^\beta)
 (t)A^{n+1}y\Big]\\
=&\sum_{j=1}^ng_{\alpha j+1-\alpha}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}
 \ast \mathbb S_\alpha^\beta)(t)A^{n+1}x\\
&+\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^{j+1}y+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast \mathbb S_\alpha^\beta)(t)A^{n+1}y\\
=&\sum_{j=0}^ng_{\alpha j+1}(t)A^{j+1}x+(g_{\alpha(n-\frac{\beta}{2})}
 \ast\mathbb S_\alpha^\beta)(t)A^{n+1}x\\
&+\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^{j+1}y+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast \mathbb S_\alpha^\beta)(t)A^{n+1}y\\
=&A\Big[\sum_{j=0}^ng_{\alpha j+1}(t)A^{j}x
 +(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb S_\alpha^\beta)(t)A^{n}x\Big]\\
&+A\Big[\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^{j}y+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast \mathbb S_\alpha^\beta)(t)A^{n}y\Big]\\
=&A \Big[D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t)x
 + D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t)y\Big]
=Au(t)
\end{align*}
and this completes the proof of the existence part in the
assertion (a).

(b) Let $x,y\in D(A^{n})$ and set
$u(t):=D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta
(t)x+D_t^{\frac{\alpha\beta}{2}}g_1\ast\mathbb S_\alpha^\beta
(t)y$. Using \eqref{eqD} in the proof of Lemma \ref{lem-reg} we
get that for every $t\ge 0$,
\begin{equation} \label{eqD2}
\begin{aligned}
u(t)=&\frac{d^k}{dt^k}\left[(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)x\right]+\frac{d^k}{dt^k}\left[(g_{k-\frac{\alpha\beta}{2}}\ast g_1\ast \mathbb S_\alpha^\beta)(t)y\right] \\
=&\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb S_\alpha^\beta)(t)A^nx \\
&+\sum_{j=0}^{n-1}g_{\alpha
j+2}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}\ast g_1\ast\mathbb
S_\alpha^\beta)(t)A^ny.
\end{aligned}
\end{equation}
It follows from \eqref{eqD2} and Lemma \ref{lem-reg} that $u\in
C([0,\infty);X)$. Using \eqref{eqD2} we get that for every $t\ge
0$,
\begin{equation} \label{V1}
\begin{aligned}
I_t^\alpha u(t):=&(g_\alpha\ast u)(t) \\
=&\sum_{j=0}^{n-1}g_{\alpha j+1+\alpha}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast \mathbb S_\alpha^\beta)(t)A^nx \\
&+\sum_{j=0}^{n-1}g_{\alpha
j+2+\alpha}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}\ast g_1\ast
g_\alpha\ast \mathbb S_\alpha^\beta)(t)A^ny.
\end{aligned}
\end{equation}
It follows from \eqref{V1} and Lemma \ref{le2.3-2}  that
$I_t^\alpha u(t)\in D(A)$ for every $t\ge 0$. Using Lemma
\ref{lem-reg}, Lemma \ref{le2.3-2} and \eqref{eqDD}, we have that
for every $t\ge 0$,
\begin{align}
u(t)=&\sum_{j=0}^{n-1}g_{\alpha
j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb
S_\alpha^\beta)(t)A^nx \nonumber \\
& +\sum_{j=0}^{n-1}g_{\alpha j+2}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast\mathbb S_\alpha^\beta)(t)A^ny \nonumber  \\
=&x+ty+\sum_{j=1}^{n-1}g_{\alpha
j+1}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb
S_\alpha^\beta)(t)A^nx \nonumber \\
&+\sum_{j=1}^{n-1}g_{\alpha j+2}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast\mathbb S_\alpha^\beta)(t)A^ny \nonumber \\
=&x+ty+A\Big[\sum_{j=1}^{n-1}g_{\alpha j+1}(t)A^{j-1}x
 +(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb S_\alpha^\beta)(t)A^{n-1}x\Big] \nonumber \\
&+A\Big[\sum_{j=1}^{n-1}g_{\alpha j+2}(t)A^{j-1}y+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast\mathbb S_\alpha^\beta)(t)A^{n-1}y\Big] \nonumber \\
=&x+ty+A\Big[\sum_{j=1}^{n}g_{\alpha j+1}(t)A^{j-1}x
 +(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast \mathbb S_\alpha^\beta)(t)A^{n}x
 \big] \nonumber  \\
&+A\Big[\sum_{j=1}^{n}g_{\alpha j+2}(t)A^{j-1}y+(g_{\alpha(n-\frac{\beta}{2})}
 \ast g_1\ast g_\alpha\ast\mathbb S_\alpha^\beta)(t)A^{n}y\Big] \nonumber \\
=&x+ty+A\Big[\sum_{j=0}^{n-1}g_{\alpha j+1+\alpha}(t)A^{j}x
 +(g_{\alpha(n-\frac{\beta}{2})}\ast g_\alpha\ast \mathbb S_\alpha^\beta)(t)A^{n}
 x\Big] \nonumber \\
&+A\Big[\sum_{j=0}^{n-1}g_{\alpha j+2+\alpha}(t)A^{j}y
 +(g_{\alpha(n-\frac{\beta}{2})}\ast g_1\ast g_\alpha\ast\mathbb S_\alpha^\beta)
 (t)A^{n}y\Big] \nonumber \\
=&x+ty+Ag_\alpha\ast \Big[\sum_{j=0}^{n-1}g_{\alpha j+1}(t)A^{j}x
 +(g_{\alpha(n-\frac{\beta}{2})}\ast \mathbb S_\alpha^\beta)(t)A^{n}x\Big] \nonumber \\
&+Ag_\alpha\ast\Big[\sum_{j=0}^{n-1}g_{\alpha j+2}(t)A^{j}y
 +(g_{\alpha(n-\frac{\beta}{2})}\ast g_1\ast \mathbb S_\alpha^\beta)(t)A^{n}y\Big] \nonumber \\
=&x+ty+A(g_\alpha\ast u)(t).   \label{e-m-s}
\end{align}
Hence, $u$ is a mild solution of \eqref{ACP-H2} and this completes
the proof of the existence part in the assertion (b).

It remains to show the uniqueness of solutions. Let $x,y\in
D(A^n)$ and let $u$ be a mild solution. We just have to show that
$(g_1\ast u)(t)$ is exponentially bounded. Using the first
equality in \eqref{e-m-s}, we have that for every $t\ge 0$,
\begin{align*}
(g_1\ast u)(t)&=\sum_{j=0}^{n-1}g_{\alpha
j+2}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})+1}\ast \mathbb
S_\alpha^\beta)(t)A^nx \\
&\quad +\sum_{j=0}^{n-1}g_{\alpha
j+3}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}\ast g_2\ast\mathbb
S_\alpha^\beta)(t)A^ny.
\end{align*}
Using Lemma \ref{lem-ext-lap} we get from the preceding equality
that there exist some constants $M,\omega\ge 0$ such that for
every $t\ge 0$,
\begin{align*}
\|(g_1\ast u)(t)\|\le Me^{\omega
t}\sum_{j=0}^n(\|A^jx\|+\|A^jy\|).
\end{align*}
We have shown that $(g_1\ast u)(t)$ is exponentially bounded. Now,
Proposition \ref{puniq-2-2} implies the uniqueness of mild and
classical solutions. The proof of the theorem is finished.
\end{proof}

\begin{remark}\rm
 We observe that although in  \eqref{ACP-H2} we have the
Caputo fractional derivative $\mathbb D_t^\alpha$, the solution is
given by the Riemann-Liouville derivative
$D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t)
x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t)
y$. If $\frac{\alpha\beta}{2}$ is not an integer, then the
function $\mathbb D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t) x+\mathbb
D_t^{\frac{\alpha\beta}{2}}(g_1\ast\mathbb  S_\alpha^\beta)(t) y$
is not a solution of  \eqref{ACP-H2}, unless $x=y=0$.
\end{remark}

\section{Resolvent families and the inhomogeneous Cauchy problem}\label{sec-inh-CP}

In this section we study the solvability and the representation of
solutions of the inhomogeneous fractional order abstract Cauchy
problem
\begin{equation}\label{ACP2}
\begin{gathered}
\mathbb D_t^\alpha u(t)=Au(t)+f(t),\quad t>0,\;1<\alpha\le 2,\\
u(0)=x,\quad  u'(0)=y,
\end{gathered}
\end{equation}
where $A$ is a closed linear operator with domain $D(A)$ defined
in a Banach space, $f:\;[0,\infty)\to X$ is a given function and
$x,y$ are given vectors in $X$.

\begin{definition} \rm
A function $u\in C([0,\infty);D(A))\cap C^1([0,\infty);X)$ is said
to be a classical solution of Problem \eqref{ACP2} if
$g_{2-\alpha}\ast\left(u-u(0)-u'(0)t\right)\in
C^2([0,\infty);X)$  and \eqref{ACP2} is satisfied.
\end{definition}

We adopt the following definition of mild solutions.

\begin{definition} \rm
A function $u\in C([0,\infty);X)$ is said to be a mild solution of
Problem \eqref{ACP2} if $I_t^\alpha u(t):=(g_\alpha\ast u)(t)\in
D(A)$  for every $t\ge 0$,  and
\begin{align*}
u(t)=x+ ty+
A\int_0^tg_\alpha(t-s)u(s)\,ds+\int_0^tg_\alpha(t-s)f(s)\,ds,\;\;\forall
t\ge 0.
\end{align*}
\end{definition}

We have the following uniqueness result.

\begin{proposition}\label{puniq-3}
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$ and let $1<\alpha\le 2$. Then the following
assertions hold.
\begin{itemize}
\item[(a)] If $u$ is a classical solution of \eqref{ACP2}, then
it is a mild solution of \eqref{ACP2}.

\item[(b)] If $(\lambda^\alpha-A)$ is invertible for $\operatorname{Re}(\lambda)$
large enough, and if a mild solution $u$ exists and $(g_1\ast u)(t)$ is
exponentially bounded, then it is unique.
\end{itemize}
\end{proposition}

\begin{proof}
Let $1<\alpha\le 2$ and let $A$ be a closed linear operator with
domain $D(A)$ defined on a Banach space $X$.

(a) Let $u$ be a classical solution of \eqref{ACP2}. Since $u\in
C([0,\infty);D(A))$, we have that $(g_\alpha\ast u)(t)\in
C([0,\infty);D(A))$. Since $\mathbb D_t^\alpha u(t)=Au(t)+f(t)$,
that is, $(g_{2-\alpha}\ast u'')(t)=Au(t)+f(t)$, we
have that $(g_\alpha\ast g_{2-\alpha}\ast
u'')(t)=A(g_\alpha\ast u)(t)+(g_\alpha\ast f)(t)$,
i.e., $(g_{2}\ast u'')(t)=A(g_\alpha\ast
u)(t)+(g_\alpha\ast f)(t)$. Hence,
$u(t)-u(0)-tu'(0)=A(g_\alpha\ast u)(t)+(g_\alpha\ast f)(t)$
for every $t\ge 0$ and we have shown that $u$ is a mild solution
of \eqref{ACP2}.

(b) Assume that \eqref{ACP2} has two mild solutions $u$ and $v$
and set $U:=u-v$. Then $U\in C([0,\infty);X)$, $(g_\alpha\ast
U)(t)\in D(A)$ for every $t\ge 0$ and $U(t)=A(g_\alpha\ast U)(t)$.
Taking the Laplace transform, we get that
$(I-\lambda^{-\alpha}A)\widehat{U}(\lambda)=0$ for
$\operatorname{Re}(\lambda)>\omega$ (where $\omega\ge 0$ is the real
number from the exponential boundedness of $(g_1\ast u)(t)$).
Since $(I-\lambda^{-\alpha}A)$ is invertible, we have that
$\widehat{U}(\lambda)=0$. By the uniqueness theorem for the
Laplace transform and by continuity, we get that $U(t)=0$ for
every $t\ge 0$. Hence, $u(t)=v(t)$ for every $t\ge 0$. The proof
is finished.
\end{proof}

\begin{remark} \rm
As for the homogeneous equation in Section \ref{sec-ho-p},
to prove the existence of mild and classical solutions of Problem
\eqref{ACP2}, we proceed by a direct method with minimal use of
 the Laplace transform.
\end{remark}

We have the following result of existence and representation of
classical and mild solutions which is the main result of this
section.

\begin{theorem}\label{th-mis}
Let $A$ be a closed linear operator with domain $D(A)$ defined on
a Banach space $X$. Let $ 1<\alpha\le 2, \beta> 0$ and set
$n:=\lceil \beta\rceil$, $k:=\lceil \frac{\alpha\beta}{2}\rceil$.
Assume that $A$ generates an $(\alpha,1)^{\beta}$-resolvent family
$\mathbb S_\alpha^{\beta}$.  Let $\mathbb P_\alpha^\beta$ be the
$(\alpha,\alpha)^{\beta}$-resolvent family generated by $A$. Then
the following assertions hold.

(a)  For every $f\in C^{k}([0,\infty);D(A))\cap C^{k+1}([0,\infty);X)$,
$f^{(2i)}(0), f^{(2i+1)}(0)\in D(A^{n+1-i})$, $i=0,1,\ldots, \frac{k-1}{2}$,
if $k$ is odd, $f^{(2i)}(0)\in D(A^{n+1-i})$, $i=0,1,\ldots, \frac k2$,
$f^{(2i+1)}(0)\in D(A^{n+1-i})$, $i=0,\ldots, \frac{k}{2}-1$, if $k$ is even,
$\mathbb D_t^{\frac{\alpha\beta}{2}}f(t):=(g_{k-\frac{\alpha\beta}{2}}
\ast f^{(k)})(t)$ is exponentially bounded, and for every $x,y\in D(A^{n+1})$,
 Problem \eqref{ACP2} has a unique classical solution $u$ given by
\begin{equation}\label{solC-2}
u(t)=D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y+D_t^{\frac{\alpha\beta}{2}} (\mathbb
P_\alpha^\beta\ast f)(t),\quad t\ge 0.
\end{equation}

(b)  For every $f\in C^{k}([0,\infty);X)$, $f^{(2i)}(0)$,
$i=0,1,\ldots, \frac{k-1}{2}$, $ f^{(2i+1)}(0)\in D(A^{n-i})$,
$i=0,1,\ldots, \frac{k-1}{2}-1$, if $k$ is odd,
$f^{(2i)}(0), f^{(2i+1)}(0)\in D(A^{n-i})$, $i=0,1,\ldots, \frac{k}{2}-1$,
if $k$ is even, $\mathbb D_t^{\frac{\alpha\beta}{2}}f(t)
:=(g_{k-\frac{\alpha\beta}{2}}\ast f^{(k)})(t)$ is exponentially bounded,
and for every $x,y\in D(A^{n})$, Problem \eqref{ACP2} has a unique mild
solution $u$ given by \eqref{solC-2}.
\end{theorem}

\begin{proof}
Let $A$, $\alpha$, $\beta$, $n$ and $k$  be as in the statement of
the theorem. First we prove existence of classical and mild
solutions.

(a)  Let $x,y\in D(A^{n+1})$. It follows from the proof of Theorem
\ref{the-m2}(a) that $D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y\in C([0,\infty);D(A))\cap C^1([0,\infty);X)$.
Moreover,
\begin{gather*}
D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(0)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(0)y=x,\\
\;\frac{d}{dt}\big[D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta\big](0)x+\frac{d}{dt}\big[D_t^{\frac{\alpha\beta}{2}}(g_1\ast
\mathbb S_\alpha^\beta)\big](0)y=y.
\end{gather*}
Now, assume that $f$ satisfies the assumptions in the statement of
part (a) of the theorem.  Using Remark \ref{rem33} and
\eqref{conv2}, we get that for every $t\ge 0$,
\begin{equation} \label{eqIP}
\begin{aligned}
&D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast f)(t)\\
&=D_t^{\frac{\alpha\beta}{2}} (g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)
 =\frac{d^k}{dt^k}\left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}
 \ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
&= \frac{d^{k-1}}{dt^{k-1}}\left[(g_{k-\frac{\alpha\beta}{2}}
 \ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta)(t) f(0)\right]
 +(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f')(t) \\
&= \sum_{i=0}^{k-1}\frac{d^{k-1-i}}{dt^{k-1-i}}\left[(g_{k-\frac{\alpha\beta}{2}}
 \ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta)(t)f^{(i)}(0)\right]
 +(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta
 \ast f^{(k)})(t) \\
&= \sum_{i=0}^{k-1}\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}
 \ast g_{\alpha-1}\ast g_{i+1} \ast \mathbb S_\alpha^\beta)(t)f^{(i)}(0)\right]
 +(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta
 \ast f^{(k)})(t) \\
&= \sum_{i=0}^{k-1}g_{\alpha}\ast\frac{d^{k}}{dt^{k}}
 \left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{i} \ast \mathbb
 S_\alpha^\beta)(t)f^{(i)}(0)\right]
 +(g_{k-\frac{\alpha\beta}{2}}\ast
 g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f^{(k)})(t).
\end{aligned}
\end{equation}

 If $k$ is odd, then using \eqref{eqIP} we have that for every $t\ge 0$,
\begin{align*}
D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast f)(t)
=&\sum_{i=0}^{\frac{k-1}{2}}g_{\alpha}\ast\frac{d^{k}}{dt^{k}}
 \left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{2i}
 \ast \mathbb S_\alpha^\beta)(t)f^{(2i)}(0)\right]\\
&+\sum_{i=0}^{\frac{k-1}{2}-1}g_{\alpha}\ast\frac{d^{k}}{dt^{k}}
 \left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{2i+1}
 \ast \mathbb S_\alpha^\beta)(t)f^{(2i+1)}(0)\right]\\
&+(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}
 \ast \mathbb S_\alpha^\beta\ast f^{(k)})(t)\\
=&\sum_{i=0}^{\frac{k-1}{2}}g_{\alpha+i(2-\alpha) }
 \ast\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}
 \ast g_{\alpha i} \ast \mathbb S_\alpha^\beta)(t)f^{(2i)}(0)\right]\\
&+\sum_{i=0}^{\frac{k-1}{2}-1}g_{\alpha+(2-\alpha)i+1}
 \ast\frac{d^{k}}{dt^{k}}\left[(g_{k-\frac{\alpha\beta}{2}}
 \ast g_{\alpha i} \ast \mathbb S_\alpha^\beta)(t)f^{(2i+1)}(0)\right]\\
&+(g_{k-\alpha\beta}\ast g_{\alpha-1}\ast \mathbb
S_\alpha^\beta\ast f^{(k)})(t).
\end{align*}
Using the preceding equality, Lemma \ref{lem-reg}(c) and Lemma
\ref{le2.3-2}, we get that for every $t\ge 0$,
\begin{align}
D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast f)(t)
=&\sum_{i=0}^{\frac{k-1}{2}}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i
 +\alpha+1}(t) A^jf^{(2i)}(0) \nonumber \\
&+ \sum_{i=0}^{\frac{k-1}{2}}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i
 +\alpha} \ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i)}(0) \nonumber\\
&+\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)
 +(2-\alpha)i+\alpha+2}(t) A^jf^{(2i+1)}(0) \nonumber\\
&+ \sum_{i=0}^{\frac{k-1}{2}-1}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i+\alpha+1 }
  \ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i+1)}(0) \nonumber\\
&+(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb
S_\alpha^\beta\ast f^{(k)})(t). \label{RR1}
\end{align} 
Using the above equality we get that for every $t\ge 0$,
\begin{equation} \label{RR2}
\begin{aligned}
&\frac{d}{dt}\big[D_t^{\frac{\alpha\beta}{2}} (\mathbb
P_\alpha^\beta\ast f)(t)\big] \\
&=\sum_{i=0}^{\frac{k-1}{2}}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i+\alpha}(t)
  A^jf^{(2i)}(0) \\
&\quad + \sum_{i=0}^{\frac{k-1}{2}}(g_{\alpha(n-\beta)+(2-\alpha)i+\alpha-1}
  \ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i)}(0) \\
&\quad +\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i
 +\alpha+1}(t) A^jf^{(2i+1)}(0) \\
&\quad + \sum_{i=0}^{\frac{k-1}{2}-1}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i
 +\alpha} \ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i+1)}(0) \\
&\quad +(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb
 S_\alpha^\beta)(t) f^{(k)}(0) +(g_{k-\frac{\alpha\beta}{2}}\ast
 g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f^{(k+1)})(t).
\end{aligned}
\end{equation}
Now, it follows from \eqref{RR1}, \eqref{RR2}, Lemma
\ref{lem-reg}, Lemma \ref{le2.3-2} and the hypothesis, that
$D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast f)(t) \in
C([0,\infty);D(A))\cap C^1([0,\infty);X)$.

 If $k$ is even, proceeding as for the case $k$ odd and using
Lemma \ref{lem-reg}, Lemma \ref{le2.3-2} and the hypothesis,
 we also get that $D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast f)(t)
\in C([0,\infty);D(A))\cap C^1([0,\infty);X)$.

From \eqref{RR1} and \eqref{RR2}, it is clear that
$D_t^{\alpha\beta} (\mathbb P_\alpha^\beta\ast
f)(0)=\frac{d}{dt}\big[D_t^{\frac{\alpha\beta}{2}} (\mathbb
P_\alpha^\beta\ast f)\big](0)=0$. We have shown that $u\in
C([0,\infty);D(A))\cap C^1([0,\infty);X)$, $u(0)=x$ and
$u'(0)=y$. By the proof of Theorem \ref{the-m2}(a) we have
that
\[
g_{2-\alpha}\ast \Big[D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x-x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y-ty\Big]\in C^2([0,\infty);X).
\]
  Using \eqref{RR1}, we have that if $k$ is odd, then for every $t\ge 0$,
\begin{align*}
\frac{d^2}{dt^2}\Big[g_{2-\alpha}\ast D_t^{\frac{\alpha\beta}{2}}
(\mathbb P_\alpha^\beta\ast f)(t)\Big]
=&\sum_{i=0}^{\frac{k-1}{2}}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i+1}(t)
  A^jf^{(2i)}(0)\\
&+ \sum_{i=0}^{\frac{k-1}{2}}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i}
\ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i)}(0)\\
&+\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)
 +(2-\alpha)i+2}(t) A^jf^{(2i+1)}(0)\\
&+ \sum_{i=0}^{\frac{k-1}{2}-1}(g_{\alpha(n-\frac{\beta}{2})
 +(2-\alpha)i+1 } \ast \mathbb S_\alpha^\beta)(t)A^{n-i}f^{(2i+1)}(0) \\
& +(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb S_\alpha^\beta)(t)
f^{(k)}(0) +(g_{k-\frac{\alpha\beta}{2}}\ast \mathbb
S_\alpha^\beta\ast f^{(k+1)})(t).
\end{align*}
We get a similar formula if $k$ is even. Therefore,
$(g_{2-\alpha}\ast D_t^{\frac{\alpha\beta}{2}} (\mathbb
P_\alpha^\beta\ast f))\in C^2([0,\infty);X)$ and hence,
$(g_{2-\alpha}\ast (u-u(0)-u'(0)g_2)\in C^2([0,\infty);X)$.
It also follows from the proof of Theorem \ref{the-m2}(a) that for
every $t\ge 0$,
\begin{equation} \label{w1}
\mathbb D_t^\alpha\Big[D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y\Big]
=A\Big[D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast\mathbb
S_\alpha^\beta)(t)y\Big].
\end{equation}
Using Lemma \ref{le2.3-2}, we get that for every $t\ge 0$,
\begin{equation} \label{w2}
\begin{aligned}
\mathbb D_t^\alpha D_t^{\frac{\alpha\beta}{2}}
(\mathbb P_\alpha^\beta\ast f)(t)
=&\mathbb D_t^\alpha D_t^{\frac{\alpha\beta}{2}} (g_{\alpha-1}
 \ast \mathbb S_\alpha^\beta\ast f)(t) \\
=&\Big(g_{2-\alpha}\ast\frac{d^2}{dt^2}\big[D_t^{\frac{\alpha\beta}{2}}
  (g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)\big]\Big)(t) \\
=&\frac{d^{k+2}}{dt^{k+2}}\left[(g_{k+2-\frac{\alpha\beta}{2}-\alpha}
 \ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
=&\frac{d^{k+2}}{dt^{k+2}}\left[(g_{k+2}\ast f)(t)
 +(g_{k+2-\frac{\alpha\beta}{2}}\ast A g_{\alpha-1}
 \ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
=&f(t)+A\frac{d^{k}}{dt^{k}}\Big[(g_{k-\frac{\alpha\beta}{2}}
 \ast  g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)\Big] \\
=&f(t)+A D_t^{\frac{\alpha\beta}{2}} (\mathbb P_\alpha^\beta\ast
f)(t).
\end{aligned}
\end{equation}
It follows from \eqref{w1} and \eqref{w2} that $\mathbb D_t^\alpha
u(t)=Au(t)+f(t)$ for every $t\ge 0$. Hence, $u$ is a classical
solution of \eqref{ACP2}  and this completes the proof of the
existence part  in the assertion (a).

(b) Let $x,y\in D(A^n)$. From the proof of Theorem
\ref{the-m2}(b) we have  $D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y\in C([0,\infty); X)$ and that $I_t^\alpha
\big[D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y\big]\in D(A)$ for all $t\ge 0$. Assume that
$f$ satisfies the hypothesis in the statement of part (b) of the
theorem. Using \eqref{RR1}, Lemma \ref{le2.3}, Lemma \ref{le2.3-2}
and Lemma \ref{lem-reg}, we have that if $k$ is odd, then for
every $t\ge 0$,
\begin{equation} \label{mm}
\begin{aligned}
I_t^\alpha D_t^{\frac{\alpha\beta}{2}}\left(\mathbb P_\alpha^\beta\ast f\right)(t)=&g_\alpha\ast D_t^{\frac{\alpha\beta}{2}}\left(g_{\alpha-1}\ast\mathbb  S_\alpha^\beta\ast f\right)(t) \\
=&\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-2-i}g_{\alpha(i+j)+(2-\alpha)i+\alpha+1}(t) A^jf^{(2i)}(0) \\
&+ \sum_{i=0}^{\frac{k-1}{2}-1}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i+\alpha} \ast \mathbb S_\alpha^\beta)(t)A^{n-1-i}f^{(2i)}(0) \\
&+\sum_{i=0}^{\frac{k-1}{2}-2}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i+\alpha+2}(t) A^jf^{(2i+1)}(0) \\
&+ \sum_{i=0}^{\frac{k-1}{2}-2}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i+\alpha+1 } \ast \mathbb S_\alpha^\beta)(t)A^{n-1-i}f^{(2i+1)}(0) \\
&+(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast g_\alpha\ast
\mathbb S_\alpha^\beta\ast f^{(k)})(t)\in D(A).
\end{aligned}
\end{equation}
We get a similar formula if $k$ is even. Hence,  for every $t\ge
0$,
\[
I_t^\alpha u(t)=I_t^\alpha D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t)x
+I_t^\alpha D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t)y
+I_t^\alpha D_t^{\frac{\alpha\beta}{2}}\left(\mathbb P_\alpha^\beta\
ast f\right)(t)\in D(A).
\]
It follows from \eqref{e-m-s} in the proof of Theorem \ref{the-m2}
that for every $t\ge 0$,
\begin{equation} \label{nn}
\begin{aligned}
&D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta(t)x+D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta)(t)y \\
&= x+ty+A\Big[(g_\alpha\ast
D_t^{\frac{\alpha\beta}{2}}\mathbb
S_\alpha^\beta)(t)x+(g_\alpha\ast
D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb
S_\alpha^\beta))(t)y\Big].
\end{aligned}
\end{equation}
Proceeding as in \eqref{eqIP} and using Lemma \ref{le2.3-2}  and
\eqref{conv1}, we have that for every $t\ge 0$,
\begin{equation} \label{nn2}
\begin{aligned}
D_t^{\frac{\alpha\beta}{2}}\left(\mathbb P_\alpha^\beta\ast f\right)(t)=&\frac{d^k}{dt^k}\left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
=&\frac{d^k}{dt^k}\left[(g_k\ast g_\alpha\ast f)(t)+A(g_\alpha\ast g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
=&(g_\alpha\ast f)(t)+Ag_\alpha\ast\frac{d^k}{dt^k}\left[(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha-1}\ast \mathbb S_\alpha^\beta\ast f)(t)\right] \\
=&(g_\alpha\ast f)(t)+A(g_\alpha\ast
D_t^{\frac{\alpha\beta}{2}}\left(\mathbb P_\alpha^\beta\ast
f\right))(t).
\end{aligned}
\end{equation}
Combining \eqref{nn} and \eqref{nn2}, we get that for every $t\ge
0$,
\begin{align*}
u(t)=&D_t^{\frac{\alpha\beta}{2}}\mathbb S_\alpha^\beta(t)x
 +D_t^{\frac{\alpha\beta}{2}}(g_1\ast \mathbb S_\alpha^\beta)(t)y
 +D_t^{\frac{\alpha\beta}{2}}\left(\mathbb P_\alpha^\beta\ast f\right)(t)\\
=&x+ty +A(g_\alpha\ast u)(t)+(g_\alpha\ast f)(t).
\end{align*}
Hence, $u$ is a mild solution of Problem \eqref{ACP2}. This
completes the proof of the existence part in the assertion (b).

It remains to show the uniqueness of solutions. Let $x,y\in
D(A^n)$ and let $f$ satisfy the assumptions in part (b) of the
theorem. Let $u$ be a mild solution. Using \eqref{eqD2} and
proceeding as in \eqref{mm} we get that, if $k$ is odd, then for
every $t\ge 0$,
\begin{equation} \label{cc}
\begin{aligned}
(g_1\ast u)(t)=&\sum_{j=0}^{n-1}g_{\alpha j+2}(t)A^jx+(g_{\alpha(n-\frac{\beta}{2})+1}\ast \mathbb S_\alpha^\beta)(t)A^nx \\
&+\sum_{j=0}^{n-1}g_{\alpha j+3}(t)A^jy+(g_{\alpha(n-\frac{\beta}{2})}\ast g_2\ast\mathbb S_\alpha^\beta)(t)A^ny \\
&+\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-2-i}g_{\alpha(i+j)+(2-\alpha)i+2}(t) A^jf^{(2i)}(0) \\
&+ \sum_{i=0}^{\frac{k-1}{2}-1}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i+1} \ast \mathbb S_\alpha^\beta)(t)A^{n-1-i}f^{(2i)}(0) \\
&+\sum_{i=0}^{\frac{k-1}{2}-2}\sum_{j=0}^{n-1-i}g_{\alpha(i+j)+(2-\alpha)i+3}(t) A^jf^{(2i+1)}(0) \\
&+ \sum_{i=0}^{\frac{k-1}{2}-2}(g_{\alpha(n-\frac{\beta}{2})+(2-\alpha)i+2} \ast \mathbb S_\alpha^\beta)(t)A^{n-1-i}f^{(2i+1)}(0) \\
&+(g_{k-\frac{\alpha\beta}{2}}\ast g_{\alpha}\ast \mathbb
S_\alpha^\beta\ast f^{(k)})(t).
\end{aligned}
\end{equation}
We get a similar equality if $k$ is even. Since by assumption
$(g_1\ast\mathbb S_\alpha^\beta)(t)$ is exponentially bounded, and
that there exist some constants $M_1,\omega_1\ge 0$ such that
$\|(g_{k-\frac{\alpha\beta}{2}}\ast f^{(k)})(t)\|\le
M_1e^{\omega_1 t}$, $t\ge 0$, it follows from \eqref{cc}  that if
$k$ is odd, then there exist some constants $M,\omega\ge 0$ such
that for every $t\ge 0$,
\begin{align*}
&\|(g_1\ast u)(t)\|\\
&\le  Me^{\omega t}\Big[\sum_{j=0}^n(\|A^jx\|+\|A^jy\|)
 +\sum_{i=0}^{\frac{k-1}{2}-1}\sum_{j=0}^{n-2-i}\| A^jf^{(2i)}(0)\|\\
&\quad + \sum_{i=0}^{\frac{k-1}{2}-1}\|A^{n-1-i}f^{(2i)}(0)\|\Big]
+Me^{\omega t}\Big[\sum_{i=0}^{\frac{k-1}{2}-2}\sum_{j=0}^{n-1-i}\|A^jf^{(2i+1)}(0)\|\\
&\quad +\sum_{i=0}^{\frac{k-1}{2}-2}\|A^{n-1-i}f^{(2i+1)}(0)\|+M_1e^{\omega_1
t}\Big].
\end{align*}
We get a similar estimate if $k$ is even. We have shown that
$(g_1\ast u)(t)$ is exponentially bounded. Now, the uniqueness of
mild and classical solutions  follows from Proposition
\ref{puniq-3} and this completes the proof.
\end{proof}

\begin{remark} \label{rmk-thm7.5} \rm
Theorem \ref{th-mis} holds in the case $\beta = 0$ as follows: 
Let $A$ be a closed linear operator with domain $D(A)$ defined on a 
Banach space $X$. Let $1<\alpha \leq 2$. Assume that $A$ generates an 
$(\alpha,1)$-resolvent family $\mathbb{S}_{\alpha}$. 
Let $\mathbb{P}_{\alpha}$ be the $(\alpha,\alpha)$-resolvent family 
generated by $A$. Then the following assertions hold.

(a) For every $f \in  C([0,\infty);D(A))\cap C^1([0,\infty);X)$  
exponentially bounded, and for every $x,y\in D(A)$,
 Problem \eqref{ACP2} has a unique classical solution 
$$
u(t)= \mathbb{S}_{\alpha}(t)x + (g_1*\mathbb{S}_{\alpha})(t)y 
+ (\mathbb{P}_{\alpha}*f)(t), \quad t\geq 0.
$$

(b) For every $f \in C([0,\infty);X)$ exponentially bounded, and 
for every $x,y\in X$, Problem \eqref{ACP2} has a unique mild solution $u$ 
given by \eqref{solC-2}.
\end{remark}


\section{Applications}\label{sec-app}

In this section we give some examples where the situations of the previous
sections are applied.
Throughout this section $\Omega\subset\mathbb{R}^N$ denotes an open set
with Lipschitz continuous boundary $\partial\Omega$. Let the real valued
coefficients satisfy $a_{ij}\in L^\infty(\Omega)$, $b_j,c_j, d\in
L^{\infty}(\Omega)$, $i,j=1,2,\ldots,N$. We assume also that there
exists a constant $\mu>0$ such that
\[
\sum_{i,j=1}^Na_{ij}(x)\xi_i\xi_j\ge \mu|\xi|^2\quad \text{for all }\xi\in\mathbb{R}^N,
\]
for a.e. $x\in\Omega$. Let $A$ be the elliptic operator formally
given by
\begin{equation}\label{eq-A}
Au=\sum_{j=1}^ND_j\Big(\sum_{i=1}^Na_{i,j}D_iu+b_ju\Big)
-\Big(\sum_{i=1}^Nc_iD_iu+du\Big).
\end{equation}

Next we have an example for Dirichlet, Neumann and Robin boundary conditions 
on $L^2$-spaces.

\begin{example} \label{ex1} \rm
For $1<\alpha\le 2$, we consider the fractional order Cauchy
problem
\begin{equation}\label{icp}
\begin{gathered}
\mathbb D_t^\alpha u(t,x)-Au(t,x)=f(t,x),\quad t>0,\;x\in\Omega,\\
\frac{\partial u(t,z)}{\partial\nu_A}+\gamma(z) u(t,z)=0,\quad t>0,\;z\in\partial\Omega,\\
 u(0,x)=u_0(x),\quad \frac{\partial u(0,x)}{\partial
t}=u_1(x),\quad x\in\Omega.
\end{gathered}
\end{equation}
Here, $u_0,u_1\in L^2(\Omega)$, $f\in C([0,\infty); L^2(\Omega))$, $A$
is the operator given in \eqref{eq-A},
\[
\frac{\partial u}{\partial\nu_A}
=\sum_{j=1}^N\Big(\sum_{i=1}^Na_{ij}D_iu+b_ju\Big)\cdot\nu_j,
\]
where $\nu$ denotes the outer normal vector of $\Omega$ at the
boundary $\partial\Omega$ and $\gamma\ge 0$ belongs to $L^\infty(\partial\Omega)$ of
$\gamma=\infty$. If $\gamma=\infty$, then the boundary conditions
in \eqref{icp} become the Dirichlet boundary conditions
$u(t,z)=0$, $t> 0$ and $z\in\partial\Omega$ (see e.g. \cite{AW1,AW2}).

We consider the first order Sobolev spaces
\[
H^1(\Omega):=\{ u\in L^2(\Omega),\;\int_{\Omega}|\nabla u|^2\,dx<\infty\}
\]
endowed with the norm
\begin{align*}
\|u\|_{H^1(\Omega)}:=\Big(\int_\Omega|u|^2\,dx+\int_\Omega|\nabla
u|^2\,dx\Big)^{1/2},
\end{align*}
and $H_0^1(\Omega)=\overline{\mathcal D(\Omega)}^{H^1(\Omega)}$ where
$\mathcal D(\Omega)$ denotes the space of test functions on $\Omega$.

Let $\mathcal A_\gamma$ be the bilinear form on $L^2(\Omega)$ with
domain $H^1(\Omega)$ and given for $u,v\in H^1(\Omega)$ by
\begin{align*}
\mathcal A_\gamma(u,v):=
&\int_{\Omega}\sum_{j=1}^N\Big(\sum_{i=1}^Na_{ij}D_iu+b_ju\Big)D_jv\,dx\\
&+\int_{\Omega}\Big(\sum_{j=1}^Nc_jD_ju+du\Big)v\,dx
+\int_{\partial\Omega}\gamma uv\,d\sigma,
\end{align*}
where $\sigma$ denotes the usual Lebesgue surface measure on the
boundary $\partial\Omega$, and let $\mathcal A_{D}$ be the bilinear form on
$L^2(\Omega)$ with domain $H_0^1(\Omega)$ and given for $u,v\in
H_0^1(\Omega)$ by
\[
\mathcal A_{D}(u,v):=
\int_{\Omega}\sum_{j=1}^N\Big(\sum_{i=1}^Na_{ij}D_iu+b_ju\Big)D_jv\,dx
+\int_{\Omega}\Big(\sum_{j=1}^Nc_jD_ju+du\big)v\,dx.
\]
It is easy to see that the bilinear forms $\mathcal A_\gamma$ and
$\mathcal A_{D}$ are closed in $L^2(\Omega)$. Let $A_{2,\gamma}$ and
$A_{2,D}$ be the closed linear operators in $L^2(\Omega)$
associated with the form $\mathcal A_\gamma$ and $\mathcal A_{D}$,
respectively. That is,
\begin{equation*}
\begin{gathered}
D(A_{2,\gamma}):=\{ u\in H^1(\Omega),\;\exists \;v\in L^2(\Omega),\;
\mathcal A_\gamma(u,\varphi)=(v,\varphi)_{L^2(\Omega)}, \;\forall
\varphi\in H^1(\Omega)\}\\
A_{2,\gamma}u=v
\end{gathered}
\end{equation*}
and
\begin{equation*}
\begin{gathered}
D(A_{2,D}):=\{ u\in H_0^1(\Omega),\;\exists \;v\in L^2(\Omega),\;
\mathcal A_{D}(u,\varphi)=(v,\varphi)_{L^2(\Omega)},
\;\forall \varphi\in H_0^1(\Omega)\}\\
A_{2,D}u=v.
\end{gathered}
\end{equation*}
One has the following more explicit description of the operators
$A_{2,\gamma}$ and $A_{2,D}$ on $L^2(\Omega)$.
\begin{gather*}
D(A_{2,\gamma})=\{u\in H^1(\Omega),\; Au\in L^2(\Omega),\;
\frac{\partial u}{\partial\nu_A}+\gamma u=0\},\quad  A_{2,\gamma}
u=Au, \\
D(A_{2,D})=\{u\in H_0^1(\Omega):\; Au\in L^2(\Omega)\},\quad A_{2,D} u=Au.
\end{gather*}
The operator $A_{2,\beta}$ (resp. $A_{2,D}$) is a realization of
the operator $A$ in $L^2(\Omega)$ with Robin boundary conditions
and Neumann boundary conditions if $\gamma=0$ (resp. with
Dirichlet boundary conditions).  With this setting Problem
\eqref{icp} can be rewritten as an abstract Cauchy problem in the
Hilbert space $L^2(\Omega)$,
\begin{equation*}
\begin{gathered}
\mathbb D_t^\alpha u(t)=\widetilde Au(t) +f(t), \;t\ge 0,\; 1<\alpha\le 2,\\
u(0)=u_0,\quad u_t(0),=u_1,
\end{gathered}
\end{equation*}
with $\widetilde A=A_{2,\gamma}$ or $A_{2,D}$. It is well-known
(see e.g. \cite{ABHN01}) that the operators $A_{2,\beta}$ and
$A_{2,D}$ generate cosine families on $L^2(\Omega)$ and hence
generate $(\alpha,1)$-resolvent families $\mathbb S_\alpha$ for
every $1<\alpha\le 2$. Therefore all the results in  Theorem
\ref{th-mis} hold for Problem \eqref{icp} with $n=k=0$.
\end{example}

Next, we consider the one-dimensional case.

\begin{example}[Elliptic operators in one-dimension]\rm
Let $a\in W^{1,\infty}(0,1)$ satisfy $a(x)\ge \mu_0>0$ for
some constant $\mu_0$. Let $b,c\in L^\infty(0,1)$, $1\le p<\infty$
and let $\alpha_j,\beta_j$ ($j=0,1$) be complex numbers such that
$(\alpha_j,\beta_j)\ne (0,0)$.  For $1<\alpha\le 2$, we consider
the fractional order Cauchy  problem
\begin{equation}\label{one}
\begin{gathered}
\mathbb D_t^\alpha u(t,x)= a(x)u_{xx}(t,x)+b(x)u_x(t,x)+c(x)u(t,x)+f(t,x),\quad
t>0,\;x\in (0,1),\\
 \alpha_j u_x(t,j)+\beta_j u(t,j)=0,\quad j=0,1,\; t> 0,\\
 u(0,x)=u_0(x),\;u_t(0,x)=u_1(x),\quad x\in (0,1).
\end{gathered}
\end{equation}
Let $\widetilde A_p$ be the operator defined on $L^p(0,1)$ by
\begin{gather*}
D(\widetilde A_p):=\{u\in W^{2,p}(0,1):
\alpha_ju'(j)+\beta_ju(j)=0,\;j=0,1\},\\
 A_pu=a(x)u''+b(x)u'+c(x)u.
\end{gather*}
The operator $\widetilde A_p$ is a realization of  $A$ (given by
$Au=a(x)u''+b(x)u'+c(x)u$) on $L^p(0,1)$ with
Dirichlet boundary conditions if $\alpha_j=0, \beta_j\ne 0$
($j=0,1$),  with Neumann boundary conditions if $\alpha_j\ne 0,
\beta_j=0$ ($j=0,1$) and Robin boundary conditions if $\alpha_j\ne
0, \beta_j\ne 0$ ($j=0,1$). With the same assumption on
$\alpha_j,\beta_j$, a realization $\widetilde A_\infty$ of $A$
with Dirichlet boundary condition on $C_0(0,1):=\{u\in C[0,1]:\;
u(0)=u(1)=0\}$ or with Neumann and Robin boundary conditions on
$C[0,1]$ is given by
\begin{gather*}
D(\widetilde A_\infty):=\{u\in C^2[0,1]:
\alpha_ju'(j)+\beta_ju(j)=0,\;j=0,1\},\\
 A_\infty u=a(x)u''+b(x)u'+c(x)u.
\end{gather*}
By \cite{CKW,KeWa} the operator $\widetilde A_p$ generates a
cosine family on $L^p(0,1)$ and $\widetilde A_\infty$ generates a
cosine family on $C[0,1]$ (on $C_0(0,1)$ if it is Dirichlet
boundary condition). The case of Wentzell (or dynamical) boundary
conditions on $L^p(0,1)\times\mathbb{C}$ and on $C[0,1]$ has been
investigated in \cite{AlWa,KeWa1}. Therefore, one has the same
results for Problem \eqref{one} as the ones given in Example
\ref{ex1}. More precisely, letting $X_p:=L^p(0,1)$ (or
$L^p(0,1)\times \mathbb{C}$ in the case of Wentzell boundary conditions)
if $1\le p<\infty$ and $X_\infty=C[0,1]$ (or $C_0(0,1)$ in the
case of Dirichlet boundary condition), then all the results in
Theorem \ref{th-mis} hold for Problem \eqref{one} with $n=k=0$.
\end{example}

\begin{example}[Elliptic operators on general $L^p$ spaces]\label{ex2}\rm
For simplicity we assume that $\Omega\subset\mathbb{R}^N$ ($N\ge 2$) is
bounded.  For $1<\alpha\le 2$, we consider the fractional order
Cauchy  problem
\begin{equation}\label{icpe2}
\begin{gathered}
\mathbb D_t^\alpha u(t,x)=Au(t,x)+f(t,x),\quad t>0,\;x\in\Omega,\\
\frac{\partial u(t,z)}{\partial\nu_A}+\gamma(z) u(t,z)=0,\quad t>0,\;z\in\partial\Omega,\\
 u(0,x)=u_0(x),\quad \frac{\partial u(0,x)}{\partial t}=u_1(x),\quad x\in\Omega.
\end{gathered}
\end{equation}
Here, $u_0, u_1\in L^p(\Omega)$, $f\in C([0,\infty); L^p(\Omega))$,  for
some $p\in [1,\infty)$ ($p\ne 2$), or $u_0,u_1\in C(\overline{\Omega})$, $f\in
C([0,\infty]; C(\overline{\Omega}))$ are given functions, and the operator $A$
is given in \eqref{eq-A}. Let $\widetilde A$ be the closed linear
operator in $L^2(\Omega)$ introduced in Example \ref{ex1}. Recall
that $\widetilde A=A_{2,\gamma}$  or $\widetilde A=A_{2,D}$. For
$2\le p<\infty$, we let $A_p$ denote the part of the operator
$A_{2,\gamma}$ in $L^p(\Omega)$ and for $1\le p<2$, we let $A_p$ be
the closure in $L^p(\Omega)$ of the operator $B$ defined by
\[
D(B)=\{u\in D(A_{2,\gamma})\cap L^p(\Omega),\; Au\in L^p(\Omega)\},\quad
 Bu=A_{2,\gamma}u=Au.
\]
The operator $A_{p}$ is a realization of the operator $A$ in
$L^p(\Omega)$ with Robin boundary conditions,  Neumann boundary
conditions if $\gamma=0$ and Dirichlet boundary conditions if
$\gamma=\infty$. By \cite{KeWa,Ouh}, the operator $A_{p}$
generates a $\beta$-times integrated cosine family
$(C_{\beta}(t))$ on $L^p(\Omega)$ with $\beta:= N|\frac
12-\frac 1p|$. Hence,  all the results in  Theorem
\ref{th-mis} hold for Problem \eqref{icpe2} with
$n:=\lceil\beta\rceil$ and $k:=\lceil\frac{\alpha\beta}{2}\rceil$.

Letting $A_\infty$ be a realization of the operator $A$ with
Robin, Neumann or Dirichlet boundary conditions on
$L^\infty(\Omega)$, we have that $A_\infty$ generates a $\beta$-times
integrated cosine family on $L^\infty(\Omega)$ with
$\beta>\frac{N}{2}$ and one can also apply Theorem \ref{th-mis}.
We notice that $D(A_\infty)$ is not dense in $L^\infty(\Omega)$.
\end{example}

Next, we consider the case of the Laplace operator on some special
open subsets of $\mathbb{R}^N$.

\begin{example}[Laplace operator on some special open sets]\rm
Let $\Omega:=\mathbb{R}^N$ or $\Omega:=(0,1)^N\subset\mathbb{R}^N$ and let
$\widetilde A_p$ be a realization of the Laplace operator on
$L^p(\Omega)$ ($p\ne 2$) with Dirichlet, Neumann or Robin boundary
conditions defined above. By \cite{ElKe,Hie,KeWa} the operator
$\widetilde A_p$ generates a $\beta$-times integrated cosine
family on $L^p(\Omega)$ with $\beta=(N-1)|\frac 12-\frac
1p|$. Therefore, one has the same results as in Example
\ref{ex2} with here $\beta=(N-1)|\frac 12-\frac 1p|$.

As in the previous example, here also, letting $A_\infty$ be a
realization of the Laplace operator with Robin, Neumann or
Dirichlet boundary conditions on $L^\infty(\Omega)$, we have that
$A_\infty$ generates a $\beta$-times integrated cosine family on
$L^\infty(\Omega)$ with $\beta=\frac{N-1}{2}$ and one can also apply
Theorem \ref{th-mis}. We also notice that $D(A_\infty)$ is not
dense in $L^\infty(\Omega)$.
\end{example}

We conclude the paper with an example involving a  Schr\"odinger
like operator.

\begin{example}\rm
We consider the fractional order Schr\"odinger like equation
\begin{equation}\label{scheq}
\begin{gathered}
\mathbb D_t^\alpha u(t,x)=e^{i\theta}\Delta_pu(t,x)+f(t,x),\quad
t> 0,\; x\in\mathbb{R}^N,\;1<\alpha<2,\\
u(0,x)=u_0(x),\quad \frac{\partial u(0,x)}{\partial t}=u_1(x),\quad
x\in\mathbb{R}^N.
\end{gathered}
\end{equation}
Here, the operator $\Delta_p$ is a realization of the Laplace
operator on $L^p(\mathbb{R}^N)$, $1\le p<\infty$, the angle $\theta$
satisfies  $\frac{\pi}{2}<\theta<
\left(1-\frac{\alpha}{4}\right)\pi$. Let
$A_p:=e^{i\theta}\Delta_p$. Then $D(A_p)=W^{2,p}(\mathbb{R}^N)$. We have
shown in Example \ref{ex-ee} that $A_p$ generates an
$(\alpha,1)=(\alpha,1)^0$-resolvent family $\mathbb S_\alpha$ on
$L^p(\mathbb{R}^N)$. Using Theorem \ref{th-mis} we get the following
results on existence of solutions to Problem \eqref{scheq}.
\begin{itemize}
\item  For every $f\in C([0,\infty);W^{2,p}(\mathbb{R}^N))\cap C^{1}([0,\infty);L^p(\mathbb{R}^N))$
  and $u_0,u_1\in W^{2,p}(\mathbb{R}^N)$, Problem \eqref{scheq} has a  classical solution
$u$.

\item  For every $f\in C([0,\infty);L^p(\mathbb{R}^N))$ and $u_0,u_1\in L^p(\mathbb{R}^N)$,
Problem \eqref{scheq} has a  mild solution $u$.
\end{itemize}
\end{example}

\subsection*{Acknowledgments}

V. Keyantuo and M. Warma are partially supported by the Air Force 
Office of Scientific Research under the Award No [FA9550-15-1-0027].
C. Lizama is partially supported by CONICYT - PIA - Anillo ACT1416
and FONDECYT grant 1140258.


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