First of all, we present some fundamental facts on the fractional calculus theory which we will use in the next section.
Definition 2.1
([1–3])
The Riemann-Liouville fractional integral of order \(\nu>0\) of a function \(h:(0, \infty)\rightarrow\mathbb{R}\) is given by
$$ I^{\nu}_{0+}h(t)=D^{-\nu}_{0+}h(t)= \frac{1}{\Gamma(\nu)} \int ^{t}_{0}(t-s)^{\nu-1}h(s)\,ds, $$
(3)
provided that the right-hand side is pointwise defined on \((0, \infty)\).
Definition 2.2
([1–3])
The Caputo fractional derivative of order \(\nu>0\) of a continuous function \(h:(0, \infty)\rightarrow\mathbb{R}\) is given by
$$ {}^{C}D^{\nu}_{0+}h(t)=\frac{1}{\Gamma(n-\nu)} \int^{t}_{0}(t-s)^{n-\nu-1} h^{n}(s)\,ds, $$
(4)
where \(n=[\nu]+1\), provided that the right-hand side is pointwise defined on \((0, \infty)\).
Lemma 2.1
([1, 3])
Assume that
\({}^{C}D^{\nu}_{0+}h(t) \in L^{1}(0,+\infty)\), \(\nu>0\). Then we have
$$ I^{\nu}_{0+} {{}^{C}D^{\nu}_{0+}}h(t)=h(t)+C_{1}+C_{2}t+ \cdots+C_{N}t^{N-1}, \quad t>0, $$
(5)
for some
\(C_{i} \in\mathbb{R}\), \(i=1, 2,\ldots,N\),where
N
is the smallest integer greater than or equal to
ν.
If h is an abstract function with values in the Banach space E, then the integrals appearing in Definition 2.1, Definition 2.2 and Lemma 2.1 are taken in Bochner’s sense. And a measurable function h is Bochner integrable if the norm of h is Lebesgue integrable.
Now let us recall some definitions and standard facts about the cone.
Let P be a cone in the ordered Banach space E, which defines a partial order on E by \(x\leq y\) if and only if \(y-x \in P\). P is normal if there exists a positive constant N such that \(\theta\leq x \leq y\) implies \(\Vert x \Vert \leq N \Vert y \Vert \), where θ is the zero element of the Banach space E. The infimum of all N with the property above is called the normal constant of P. For more details on the cone P, we refer the readers to [28, 29].
Throughout the paper, we set E be an ordered Banach space with the norm \(\Vert \cdot \Vert \) and the partial order ‘≤’. Let \(P=\{x\in E\vert x\geq\theta\}\) be a positive cone, which is normal with normal constant N. Let \(J=[0,+\infty)\). Set
$$BC(J,E)= \bigl\{ u(t)\vert u(t) \text{is continuous and bounded on }J \bigr\} . $$
Obviously, \(BC(J,E)\) is a Banach space with the norm \(\Vert u \Vert _{{b}}=\sup_{{t\in J}} \Vert u(t) \Vert \). Let
$$P_{B}= \bigl\{ u\in BC(J,E)\vert u(t)\geq\theta, t\in J \bigr\} . $$
It is easy to see that \(P_{B}\) is also normal with the same normal constant N of the cone P. Besides, \(BC(J,E)\) is also an ordered Banach space with the partial order ‘≤’ induced by the positive cone \(P_{B}\) (without confusion, we denote by ‘≤’ the partial order on both E and \(BC(J,E)\)).
We denote by \([v,w]\) the order interval \(\{u\in P_{B}\vert v\leq u\leq w, v, w\in BC(J,E)\}\) on \(BC(J,E)\), and use \([v(t),w(t)]\) to denote the order interval \(\{z\in E\vert v(t)\leq z\leq w(t)\}\) on E for \(t\in J\).
Next, we give some facts about the semigroups of linear operators. These results can be found in [30, 31].
For a strongly continuous semigroup (i.e., \(C_{0}\)-semigroup) \(\{T(t)\} _{t\geq0}\), the infinitesimal generator of \(\{T(t)\}_{{t\geq0}}\) is defined by
$$Ax=\lim_{t\rightarrow0^{+}}\frac{T(t)x-x}{t},\quad x\in E. $$
We denote by \(D(A)\) the domain of A, that is,
$$D(A)= \biggl\{ x\in E \Big\vert \lim_{t\rightarrow0^{+}}\frac{T(t)x-x}{t} \text{ exists} \biggr\} . $$
Lemma 2.2
([30, 31])
Let
\(\{T(t)\}_{t\geq0}\)
be a
\(C_{0}\)-semigroup, then there exist constants
\(C\geq1\)
and
\(\omega\in\mathbb{R}\)
such that
\(\Vert T(t) \Vert \leq C e^{\omega t}\), \(t\geq0\).
Lemma 2.3
([30, 31])
A linear operator
A
is the infinitesimal generator of a
\(C_{0}\)-semigroup
\(\{T(t)\}_{t\geq0}\)
if and only if
-
(i)
A
is closed and
\(\overline{D(A)}=E\).
-
(ii)
The resolvent set
\(\rho(A)\)
of
A
contains
\(\mathbb{R}^{+}\)
and, for every
\(\lambda>0\), we have
$$\bigl\Vert R(\lambda,A) \bigr\Vert \leq\frac{1}{\lambda}, $$
where
$$R(\lambda,A):=(\lambda I-A)^{-1}= \int_{0}^{+\infty}e^{-\lambda t}T(t)x\,dt,\quad x\in E. $$
Definition 2.3
([30, 31])
A \(C_{0}\)-semigroup \(\{T(t)\}_{{t\geq0}}\) is said to be uniformly exponentially stable if \(\omega_{0}<0\), where \(\omega_{0}\) is the growth bound of \(\{T(t)\}_{{t\geq0}}\), which is defined by
$$\omega_{0}=\inf\bigl\{ \omega\in\mathbb{R}\vert\exists C\geq1 \text{ such that } \bigl\Vert T(t) \bigr\Vert \leq C e^{\omega t},t\geq0 \bigr\} . $$
Definition 2.4
([27])
A \(C_{0}\)-semigroup \(\{T(t)\}_{{t\geq0}}\) is said to be positive on E, if order inequality \(T(t)x\geq\theta\), \(x\in E\) and \(t\geq0\).
According to Lemma 2.2 and Definition 2.3, if \(\{ T(t)\}_{{t\geq0}}\) is a uniformly exponentially stable \(C_{0}\)-semigroup with the growth bound \(\omega_{0}\), then for any \(\omega\in(0, \vert \omega_{0} \vert ]\), there exists a constant \(C\geq1\) such that \(\Vert T(t) \Vert \leq C e^{\omega t}\), \(t\geq0\). Now, we define a norm in E by
$$\Vert x \Vert _{\omega}=\sup_{t\geq0} \bigl\Vert e^{\omega t}T(t)x \bigr\Vert . $$
Evidently, we have \(\Vert x \Vert \leq \Vert x \Vert _{\omega}\leq C \Vert x \Vert \), that is to say, the norm \(\Vert \cdot \Vert _{\omega}\) and \(\Vert \cdot \Vert \) are equivalent. We denote by \(\Vert T(t) \Vert _{\omega}\) the norm of \(T(t)\) induced by the norm \(\Vert \cdot \Vert _{\omega}\). Then we obtain
$$ \bigl\Vert T(t) \bigr\Vert _{\omega}\leq e^{-\omega t},\quad t\geq0. $$
(6)
Also, we can define the equivalent norm on \(BC(J,E)\) by
$$\Vert u \Vert _{{b\omega}}=\sup_{{t\in J}} \bigl\Vert u(t) \bigr\Vert _{{\omega}}, \quad u\in BC(J,E). $$
Obviously, if \(u(t)\equiv u_{0}\), \(t\in J\), \(u_{0}\in E\), then we have
$$\Vert u \Vert _{{b\omega}}= \Vert u_{0} \Vert _{{b\omega}}= \Vert u \Vert _{{\omega}}. $$
Consider the one-sided stable probability density [15, 16, 32]
$$\psi_{q}(\theta)=\frac{1}{\pi}\sum_{n=1}^{\infty} \theta^{-q n-1}\frac {\Gamma(nq+1)}{n!}\sin(n\pi q),\quad\theta\in(0,\infty), $$
where \(0< q<1\). By Remark 2.8 in [16], for \(0\leq\gamma\leq1\), one has
$$ \int_{0}^{\infty}\theta^{-q\gamma}\psi_{q}( \theta)\,d\theta=\frac{\Gamma ({1+\gamma})}{\Gamma({1+q\gamma})}. $$
(7)
From [15, 16, 32], the Laplace transform of the one-sided stable probability density \(\psi_{q}(\theta)\) is given by
$$ \mathcal{L}\bigl[\psi_{q}(\theta)\bigr]= \int_{0}^{\infty}e^{-\lambda\theta}\psi _{q}( \theta)\,d\theta=e^{{-{\lambda}^{q}}}, \quad0< q< 1. $$
(8)
In the following, we assume that \(\{T(t)\}_{{t\geq0}}\) is a uniformly exponentially stable \(C_{0}\)-semigroup with the growth bound \(\omega_{0}\), and \(\omega\in(0, \vert \omega_{0} \vert )\).
Lemma 2.4
Define an operator
$$\begin{aligned}& (\Upsilon h) (t): = \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}\frac{\psi_{q}(\theta )}{\theta^{q}} \tau^{-q}(1-\tau)^{q-1}T \biggl(\frac{t^{q}(1-\tau )^{q}}{\theta^{q}} \biggr)h(s)\,d \theta \,d\tau, \\& \quad h\in BC(J,E). \end{aligned}$$
(9)
Then
\(\Upsilon:BC(J,E)\longrightarrow BC(J,E)\)
and
$$ \Vert \Upsilon h \Vert _{{b\omega}}\leq \Vert h \Vert _{{b\omega}}. $$
In particular, if
\(h(t)\equiv x\), \(t\in J\), \(x\in E\), then
$$ \Vert \Upsilon x \Vert _{{b\omega}}\leq \Vert x \Vert _{{_{\omega}}}. $$
Proof
Since
$$\begin{aligned}& \bigl\Vert (\Upsilon h) (t) \bigr\Vert _{{\omega}} \\& \quad \leq \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}\tau^{-q}(1- \tau )^{q-1}\frac{\psi_{q}(\theta)}{\theta^{q}} \biggl\Vert T \biggl(\frac {t^{q}(1-\tau)^{q}}{\theta^{q}} \biggr)h(s) \biggr\Vert _{{\omega }}\,d\theta \,d\tau \\& \quad \leq \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}\tau^{-q}(1- \tau )^{q-1}\frac{\psi_{q}(\theta)}{\theta^{q}} \biggl\Vert T \biggl(\frac {t^{q}(1-\tau)^{q}}{\theta^{q}} \biggr) \biggr\Vert _{{\omega}} \bigl\Vert h(s) \bigr\Vert _{{\omega}}\,d\theta \,d\tau \\& \quad \leq \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}\tau^{-q}(1- \tau )^{q-1}\frac{\psi_{q}(\theta)}{\theta^{q}}e^{-\omega (\frac {t^{q}(1-\tau)^{q}}{\theta^{q}} )} \Vert h \Vert _{{b\omega}}\,d\theta \,d\tau \\& \quad \leq \frac{q}{\Gamma{(1-q)}} \Vert h \Vert _{{b\omega}} \int _{0}^{1}\tau^{-q}(1- \tau)^{q-1} \biggl( \int_{0}^{\infty}\frac{\psi_{q}(\theta )}{\theta^{q}}\,d\theta \biggr) \,d\tau \\& \quad \leq \Vert h \Vert _{{b\omega}}, \end{aligned}$$
then the proof is finished. □
Lemma 2.5
Define a linear operator
\(\mathcal{R}:BC(J,E)\longrightarrow BC(J,E)\)
as
$$\begin{aligned} \mathcal{R}h :=&\sum_{i=1}^{\infty} \sigma_{i}\bigl[(\Upsilon h) (\tau_{i})\bigr] \\ =&\sum_{i=1}^{\infty}\sigma_{i} \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma {(1-q)}}\frac{\psi_{q}(\theta)}{\theta^{q}} \tau^{-q}(1-\tau)^{q-1}T \biggl(\tau_{i}^{q} \frac{(1-\tau)^{q}}{\theta^{q}} \biggr)x\,d\theta \,d\tau. \end{aligned}$$
Then
\(\mathcal{R}\)
is bounded and
\(\Vert \mathcal{R} \Vert _{{\omega}}\leq\sum_{i=1}^{\infty}\sigma_{i} \). Besides, if
\(0<\sum_{i=1}^{\infty}\sigma_{i}<1\), then
\((I-\mathcal{R})^{-1}\)
is a linear bounded operator and
$$\bigl\Vert (I-\mathcal{R})^{-1} \bigr\Vert _{{\omega}}\leq \frac {1}{1-\sum_{i=1}^{\infty}\sigma_{i}}. $$
Proof
In view of Lemma 2.4, one can get
$$\begin{aligned} \Vert \mathcal{R}h \Vert _{{\omega}} =& \Biggl\Vert \sum _{i=1}^{\infty}\sigma_{i}\bigl[(\Upsilon h) ( \tau_{i})\bigr] \Biggr\Vert _{{\omega}} \leq\sum _{i=1}^{\infty}\sigma_{i} \bigl\Vert \bigl[( \Upsilon h) (\tau_{i})\bigr] \bigr\Vert _{{\omega}}\leq\sum _{i=1}^{\infty}\sigma_{i} \Vert \Upsilon h \Vert _{{b\omega}} \\ \leq& \Biggl(\sum_{i=1}^{\infty} \sigma_{i} \Biggr) \Vert h \Vert _{{_{\omega}}}. \end{aligned}$$
Hence, \(\mathcal{R}\) is bounded and \(\Vert \mathcal{R} \Vert _{{\omega}}\leq\sum_{i=1}^{\infty}\sigma_{i} \). □
Lemma 2.6
Set
$$(\mathcal{W}h) (t)= \int_{0}^{ t} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}(t-s)^{q-1}T \biggl(\frac{(t-s)^{q}}{\theta^{q}} \biggr)h(s)\,d\theta \,ds,\quad h\in BC(J,E). $$
Then
\(\mathcal{W}:BC(J,E)\longrightarrow BC(J,E)\)
and
$$\bigl\Vert (\mathcal{W}h) (t) \bigr\Vert _{{\omega}} \leq \frac{1}{\omega}\frac{1}{\Gamma{(q+1)}} \Vert h \Vert _{{b\omega}};\qquad \bigl\Vert (\mathcal{W}h) \bigr\Vert _{{b\omega }}\leq \frac{1}{\omega} \frac{1}{\Gamma{(q+1)}} \Vert h \Vert _{{b\omega}}. $$
Proof
Since
$$\begin{aligned} (\mathcal{W}h) (t) =& \int_{0}^{ t} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}(t-s)^{q-1}T \biggl(\frac{(t-s)^{q}}{\theta^{q}} \biggr)h(s)\,d\theta \,ds \\ =& \int_{0}^{ 1} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}t^{q}(1- \tau )^{q-1}T \biggl(\frac{t^{q}(1-\tau)^{q}}{\theta^{q}} \biggr)h(t\tau )\,d\theta \,d\tau, \end{aligned}$$
then
$$\begin{aligned}& \bigl\Vert (\mathcal{W}h) (t) \bigr\Vert _{{\omega}} \\& \quad \leq \int_{0}^{ 1} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}t^{q}(1- \tau )^{q-1} \biggl\Vert T \biggl(\frac{t^{q}(1-\tau)^{q}}{\theta^{q}} \biggr) \biggr\Vert _{{\omega}} \bigl\Vert h(t\tau) \bigr\Vert _{{\omega }}\,d\theta \,d\tau \\& \quad \leq \int_{0}^{ 1} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}t^{q}(1- \tau )^{q-1}e^{-\omega\frac{t^{q}(1-\tau)^{q}}{\theta^{q}}} \bigl\Vert h(t\tau ) \bigr\Vert _{{\omega}}\,d\theta \,d\tau \\& \quad \leq \frac{1}{\omega} \Vert h \Vert _{{b\omega}} \int_{0}^{\infty } \biggl[ \biggl( \int_{0}^{1} e^{-\omega (\frac{t^{q}(1-\tau)^{q}}{\theta ^{q}} )}\,d \biggl(-\omega \frac{t^{q}(1-\tau)^{q}}{\theta^{q}} \biggr) \biggr)\frac{\psi_{q}(\theta)}{\theta^{q}} \biggr]\,d\theta \\& \quad = \frac{1}{\omega} \Vert h \Vert _{{b\omega}} \int_{0}^{\infty } \bigl(1-e^{-\omega\frac{t^{q}}{\theta^{q}}} \bigr) \frac{\psi_{q}(\theta )}{\theta^{q}}\,d\theta \\& \quad \leq \frac{1}{\omega}\frac{1}{\Gamma{(q+1)}} \Vert h \Vert _{{b\omega}}. \end{aligned}$$
Therefore,
$$\bigl\Vert (\mathcal{W}h) \bigr\Vert _{{b\omega}}\leq \frac{1}{\omega } \frac{1}{\Gamma{(q+1)}} \Vert h \Vert _{{b\omega}}. $$
□
Lemma 2.7
Let
\(h \in BC(J,E)\)
and
\(u_{0}\in D(A)\). Assume that
\(0<\sum_{i=1}^{\infty }\sigma_{i}<1\). Then the linear fractional evolution equation
$$\begin{aligned} \textstyle\begin{cases} {}^{C}D_{0+}^{q}u(t)=Au(t)+h(t),& t \in(0,+\infty),\\ u(0)=u_{0}, \end{cases}\displaystyle \end{aligned}$$
(10)
has a unique solution
\(u \in BC(J,E)\)
of the following form:
$$\begin{aligned} u(t) =&(\Upsilon u_{0}) (t)+(\mathcal{W}h) (t) \\ =& \int_{0}^{1} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}\frac{\psi_{q}(\theta )}{\theta^{q}} \tau^{-q}(1-\tau)^{q-1}T \biggl(\frac{t^{q}(1-\tau )^{q}}{\theta^{q}} \biggr)u_{0}\,d\theta \,d\tau \\ &{}+ \int_{0}^{ t} \int_{0}^{\infty}q\frac{\psi_{q}(\theta)}{\theta^{q}}(t-s)^{q-1}T \biggl(\frac{(t-s)^{q}}{\theta^{q}} \biggr)h(s)\,d\theta \,ds. \end{aligned}$$
(11)
Proof
In view of Definitions 2.1, 2.2 and Lemma 2.1, equation (10) can be rewritten by the equivalent integral equation as follows:
$$ u(t)=u_{0}+\frac{1}{\Gamma{(q)}} \int_{0}^{t}(t-s)^{q-1}\bigl[Au(s)+h(s) \bigr] \,ds. $$
(12)
Denote by \(U(\lambda)\) and \(H(\lambda)\) the Laplace transform of \(u(t)\) and \(h(t)\), respectively, using a similar method as that in [13, 14], then with the Laplace transform, we can rewrite the above equation as
$$ U(\lambda)=\frac{1}{\lambda}u_{0}+ \frac{1}{\lambda^{q}}AU(\lambda)+\frac {1}{\lambda^{q}}H(\lambda),\quad\lambda>0. $$
(13)
Then one has
$$\bigl(\lambda^{q}I-A\bigr)U(\lambda)=\lambda^{q-1}u_{0}+H( \lambda). $$
By virtue of (8) and Lemma 2.3, we obtain
$$\begin{aligned} U(\lambda) =&\bigl(\lambda^{q}I-A\bigr)^{-1} \lambda^{q-1}u_{0}+\bigl(\lambda ^{q}I-A \bigr)^{-1}H(\lambda) \\ =&\lambda^{q-1} \int_{0}^{\infty}e^{-\lambda^{q}s}T(s)u_{0}\,ds+ \int_{0}^{\infty }e^{-\lambda^{q}s}T(s)H(\lambda)\,ds \\ =&\lambda^{q-1} \int_{0}^{\infty} \int_{0}^{\infty}e^{-\lambda s^{1/q}\theta }\psi_{q}( \theta)T(s)u_{0}\,d\theta \,ds \\ &{}+ \int_{0}^{\infty} \int_{0}^{\infty}e^{-\lambda s^{1/q}\theta}\psi _{q}( \theta)T(s)H(\lambda)\,d\theta \,ds \\ =&\lambda^{q-1} \int_{0}^{\infty}e^{-\lambda t} \biggl[ \int_{0}^{\infty}q \frac{t^{q-1}}{\theta^{q}}\psi_{q}( \theta)T \biggl(\frac{t^{q}}{\theta ^{q}} \biggr)u_{0}\,d\theta \biggr]\,dt \\ &{}+ \int_{0}^{\infty}e^{-\lambda t} \biggl[ \int_{0}^{t} \int_{0}^{\infty}q \frac {(t-s)^{q-1}}{\theta^{q}}\psi_{q}( \theta)T \biggl(\frac{(t-s)^{q}}{\theta ^{q}} \biggr)h(s)\,d\theta \,ds \biggr]\,dt. \end{aligned}$$
By the definition of Laplace transforms and the convolution theorem, applying Lemma 2.6 and the inverse Laplace transforms on the above equations, then one can derive that
$$\begin{aligned} u(t) =&\mathcal{L}^{-1}\bigl[\lambda^{q-1}\bigr]\ast \mathcal{L}^{-1} \biggl[ \int _{0}^{\infty}e^{-\lambda t} \biggl[ \int_{0}^{\infty}q \frac{t^{q-1}}{\theta ^{q}}\psi_{q}( \theta)T \biggl(\frac{t^{q}}{\theta^{q}} \biggr)u_{0}\,d\theta \biggr]\,dt \biggr] \\ &{}+\mathcal{L}^{-1} \biggl[ \int_{0}^{\infty}e^{-\lambda t} \biggl( \int_{0}^{t} \int _{0}^{\infty}q \frac{(t-s)^{q-1}}{\theta^{q}}\psi_{q}( \theta)T \biggl(\frac {(t-s)^{q}}{\theta^{q}} \biggr)h(s)\,d\theta \,ds \biggr)\,dt \biggr] \\ =&\frac{t^{-q}}{\Gamma{(1-q)}}\ast \biggl[ \int_{0}^{\infty}q \frac {t^{q-1}}{\theta^{q}}\psi_{q}( \theta)T \biggl(\frac{t^{q}}{\theta ^{q}} \biggr)u_{0}\,d\theta \biggr] \\ &{}+ \int_{0}^{t} \int_{0}^{\infty}q \frac{(t-s)^{q-1}}{\theta^{q}}\psi _{q}( \theta)T \biggl(\frac{(t-s)^{q}}{\theta^{q}} \biggr)h(s)\,d\theta \,ds \\ =& \int_{0}^{t} \int_{0}^{\infty}\frac{q}{\Gamma{(1-q)}}s^{-q} \frac {(t-s)^{q-1}}{\theta^{q}}\psi_{q}(\theta)T \biggl(\frac{(t-s)^{q}}{\theta ^{q}} \biggr)u_{0}\,d\theta \,ds+(\mathcal{W}h) (t) \\ =&(\Upsilon u_{0}) (t)+(\mathcal{W}h) (t). \end{aligned}$$
Since
$$\bigl\Vert (\Upsilon u_{0}) (t)+(\mathcal{W}h) (t) \bigr\Vert _{{\omega }}\leq \bigl\Vert (\Upsilon u_{0}) (t) \bigr\Vert _{{\omega}}+ \bigl\Vert (\mathcal{W}h) (t) \bigr\Vert _{{\omega}}\leq \Vert u_{0} \Vert _{{\omega}}+\frac{1}{\omega} \frac{1}{\Gamma{(q+1)}} \Vert h \Vert _{{b\omega}}, $$
then
$$\Vert \Upsilon u_{0}+\mathcal{W}h \Vert _{{b\omega}}\leq \Vert u_{0} \Vert _{{\omega}}+\frac{1}{\omega} \frac{1}{\Gamma {(q+1)}} \Vert h \Vert _{{b\omega}}. $$
Hence, \(u\in BC(J,E)\). Then we complete the proof. □