According to [14], we introduce the definition of the mild solution to problem (1.1).
Definition 4.1
A continuous function \(x:[-\tau,b]\rightarrow X\) is a mild solution of problem (1.1) if \(x(t)=\phi(t)\) on \([-\tau,0]\) and there exists \(f\in L^{1}([0,b],X)\) such that \(f(t)\in F(t,x(t))\) for a.e. \(t\in[0,b]\) and
$$\begin{aligned} x(t) =&\frak{S}_{q}(t) \bigl[\phi(0)-g(0,\phi) \bigr]+g(t,x_{t})+ \int _{0}^{t}(t-s)^{q-1}A \frak{T}_{q}(t-s)g(s,x_{s})\,ds \\ &{} + \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds, \end{aligned}$$
where
$$\begin{aligned}& \frak{S}_{q}(t)= \int_{0}^{\infty}\xi_{q}(\theta)T \bigl(t^{q}\theta \bigr)\,d\theta,\qquad \frak{T}_{q}(t)=q \int_{0}^{\infty}\theta\xi_{q}(\theta)T \bigl(t^{q}\theta \bigr)\,d\theta, \\& \xi_{q}(\theta)=\frac{1}{q}\theta^{-1-\frac{1}{q}} \Psi_{q} \bigl(\theta^{-\frac{1}{q}} \bigr), \\& \Psi_{q}(\theta)=\frac{1}{\pi}\sum_{n=1}^{\infty}(-1)^{n-1} \theta^{-qn-1} \frac{\Gamma(nq+1)}{n!}\sin(n\pi q), \quad\theta\in{\mathbf {R}}^{+}. \end{aligned}$$
Remark 4.2
([25])
The function \(\xi_{q}(\theta)\) is a probability density function defined on \(\mathbf{R}^{+}\), and
$$\int_{0}^{\infty}\theta^{v} \xi_{q}(\theta)\,d\theta= \int_{0}^{\infty}\frac{1}{\theta^{qv}}\Psi_{q}( \theta)\,d\theta=\frac{\Gamma(1+v)}{\Gamma (1+qv)}. $$
Lemma 4.3
([14])
The operators
\(\frak{S}_{q}\)
and
\(\frak{T}_{q}\)
have the following properties:
-
(i)
For any fixed
\(t\geq0\), \(\frak{S}_{q}\)
and
\(\frak{T}_{q}\)
are linear and bounded operators, that is, for any
\(x\in X\),
$$\bigl|\frak{S}_{q}(t)x \bigr|\leq M|x|, \qquad \bigl|\frak{T}_{q}(t)x \bigr| \leq \frac{qM}{\Gamma(1+q)}|x|. $$
-
(ii)
\(\{\frak{S}_{q}(t)\}_{t\geq0}\)
and
\(\{\frak{T}_{q}(t)\}_{t\geq 0}\)
are strongly continuous.
-
(iii)
If
\(T(t)\)
is compact for
\(t>0\), then
\(\frak{S}_{q}(t)\)
and
\(\frak{T}_{q}(t)\)
are also compact operators for
\(t>0\).
-
(iv)
For any
\(x\in X\), \(\beta\in(0,1)\), and
\(\eta\in(0,1]\), we have
$$A\frak{T}_{q}(t)x=A^{1-\beta}\frak{T}_{q}(t)A^{\beta}x, \quad 0\leq t\leq b, $$
and
$$\bigl\| A^{\eta}\frak{T}_{q}(t) \bigr\| \leq\frac{qC_{\eta}\Gamma(2-\eta)}{t^{q\eta}\Gamma(1+q(1-\eta))},\quad 0< t \leq b. $$
Let us list the following hypotheses:
(H1) \(T(t)\) is compact operator for every \(t>0\).
(H2) The multivalued map \(F:[0,b]\times C([-\tau,0],X)\rightarrow {\mathcal{P}}_{b,cl,cv}(X)\) satisfies the following conditions:
-
(i)
For each \(t\in[0,b]\), \(F(t,\cdot)\) is upper semicontinuous, for each \(x \in C([-\tau,0],X)\), \(F(\cdot,x)\) is measurable, and the set \(N_{F,x}=\{f\in L^{1}([0,b],X): f(t)\in F(t,x), \mbox{for a.e. } t\in [0,b]\}\) is not empty.
-
(ii)
For each \(x\in C([-\tau,0],X)\), there exist \(m\in L^{1/q_{1}}([0,b],{\mathbf {R}}^{+})\) and \(r\in C([0,b],{\mathbf {R}}^{+})\) such that
$$\sup \bigl\{ \bigl|f(t) \bigr|: f(t)\in F(t, x) \bigr\} \leq m(t)+r(t)\|x \|_{[-\tau,0]} \quad \mbox{for a.e. } t\in [0,b], $$
where \(q_{1}\in[0,q)\).
(H3) There exists a constant \(\beta\in(0,1)\) such that \(g\in D(A^{\beta})\), \(A^{\beta}g\) is continuous, and
-
(i)
there exists a positive constant L such that
$$\bigl|A^{\beta}g(t_{1},x_{1})-A^{\beta}g(t_{2}, x_{2}) \bigr|\leq L \bigl(|t_{1}-t_{2}|+ \|x_{1}-x_{2}\| _{[-\tau,0]} \bigr) $$
for \(0< t_{1},t_{2}< b\), \(x_{1},x_{2}\in C([-\tau,0],X)\), and
-
(ii)
there exist positive constants \(L_{1}\), \(L_{2}\) such that
$$\bigl|A^{\beta}g(t,x) \bigr|\leq L_{1}\|x\|_{[-\tau,0]}+L_{2} $$
for any \(x\in C([-\tau,0],X)\).
Theorem 4.4
Assume that hypotheses (H1)-(H3) are satisfied. Then problem (1.1) has one mild solution, provided that
$$ L_{0}=L \biggl( \bigl\| A^{-\beta} \bigr\| +\frac{b^{q\beta}C_{1-\beta}\Gamma(1+\beta )}{\beta\Gamma(1+q\beta)} \biggr)< 1, \quad \bigl\| A^{-\beta} \bigr\| L_{1}< 1. $$
(4.1)
Proof
Transform problem (1.1) into a fixed point problem. Consider the multivalued operator \(\Phi:C([-\tau ,b],X)\rightarrow { \mathcal {P}}(C([-\tau,b],X))\) where Φx is defined as the set of \(\rho\in C([-\tau,b],X)\) such that
$$\rho(t)= \textstyle\begin{cases} \phi(t), & t\in[-\tau,0],\\ \frak{S}_{q}(t)[\phi(0)-g(0,\phi)]+g(t,x_{t}) +\int_{0}^{t}(t-s)^{q-1}A\frak{T}_{q}(t-s)g(s,x_{s})\,ds\\ \quad{}+\int_{0}^{t}(t-s)^{q-1}\frak{T}_{q}(t-s)f(s)\,ds, & t\in[0,b], \end{cases} $$
where \(f\in N_{F,x}=\{f\in L^{1}([0,b], X):f(t)\in F(t,x_{t}) \mbox{ for a.e. } t\in[0,b]\}\). We know that the fixed point of Φ is the mild solution of problem (1.1). Consider the operators \({\mathcal {A}}: C([-\tau,b],X)\rightarrow C([-\tau,b],X)\) and \({\mathcal {B}}:C([-\tau,b],X) \rightarrow { \mathcal {P}}(C([-\tau,b],X))\) defined by
$$({\mathcal {A}}x) (t)= \textstyle\begin{cases} \phi(t), & t\in[-\tau,0],\\ \frak{S}_{q}(t)[\phi(0)-g(0,\phi)]+g(t,x_{t})\\ \quad{}+\int_{0}^{t}(t-s)^{q-1}A\frak {T}_{q}(t-s)g(s,x_{s})\,ds, & t\in[0,b], \end{cases} $$
and
$${\mathcal {B}}x= \left \{\rho\in C \bigl([-\tau,b],X \bigr): \rho(t)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0, & t\in[-\tau,0],\\ \int_{0}^{t}(t-s)^{q-1}\frak{T}_{q}(t-s)f(s)\,ds,& f\in N_{F,x}, t\in[0,b] \end{array}\displaystyle \right . \right \}. $$
It is clear that \(\Phi={\mathcal {A}}+{\mathcal {B}}\). Mild solutions of problem (1.1) are converted to the fixed points of \(x\in{\mathcal {A}}x+{\mathcal {B}}x \). We shall show that the operators \({\mathcal {A}}\) and \({\mathcal {B}}\) satisfy the conditions of Theorem 2.7 by the following steps.
Step 1. \({\mathcal {A}}\) is a contraction.
Let \(x, y\in C([-\tau,b],X)\). Then, for each \(t\in[0,b]\), from condition (H3) we have that
$$\begin{aligned} \bigl|({\mathcal {A}}x) (t)-({\mathcal {A}}y) (t)\bigr| =& \biggl|g(t,x_{t})+ \int _{0}^{t}(t-s)^{q-1}A \frak{T}_{q}(t-s)g(s,x_{s})\,ds \\ &{}-g(t,y_{t})- \int_{0}^{t}(t-s)^{q-1}A \frak{T}_{q}(t-s)g(s,y_{s})\,ds \biggr| \\ \leq&\bigl|g(t,x_{t})-g(t,y_{t})\bigr|+ \biggl| \int_{0}^{t}(t-s)^{q-1}A\frak {T}_{q}(t-s) \bigl(g(s,x_{s})-g(t,y_{s}) \bigr) \,ds \biggr| \\ \leq& L \bigl\| A^{-\beta}\bigr\| \|x_{t}-y_{t} \|_{[-\tau,0]}\\ &{}+L \int_{0}^{t}(t-s)^{q-1}\bigl\| A^{1-\beta} \frak {T}_{q}(t-s)\bigr\| \|x_{s}-y_{s} \|_{[-\tau,0]}\,ds \\ \leq& L \biggl(\bigl\| A^{-\beta}\bigr\| +\frac{qC_{1-\beta}\Gamma(1+\beta)}{\Gamma (1+q\beta)} \int_{0}^{t}(t-s)^{q\beta-1}\,ds \biggr) \sup _{0\leq s\leq t}\|x_{s}-y_{s}\|_{[-\tau,0]} \\ \leq& L \biggl(\bigl\| A^{-\beta}\bigr\| +\frac{b^{q\beta}C_{1-\beta}\Gamma(1+\beta )}{\beta\Gamma(1+q\beta)} \biggr) \|x-y \|_{[-\tau,b]} \\ =&L_{0}\|x-y\|_{[-\tau,b]}. \end{aligned}$$
Therefore,
$$\|{\mathcal {A}}x-{\mathcal {A}}y\|_{[-\tau,b]}\leq L_{0}\|x-y \|_{[-\tau,b]}, $$
so that \({\mathcal {A}}\) is a contraction since \(L_{0}<1\).
Step 2. \({\mathcal {B}}x\) is convex for each \(x\in C([-\tau,b],X)\).
Indeed, if \(\rho_{1}\) and \(\rho_{2}\) belong to \({\mathcal {B}}x\), then there exist \(f_{1}, f_{2}\in N_{F,x}\) such that for each \(t\in [0,b]\), we have
$$\begin{aligned} \rho_{i}(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f_{i}(s)\,ds, \quad i=1, 2. \end{aligned}$$
Let \(0\leq\lambda\leq1\). Then for each \(t\in[0,b]\), we have
$$\begin{aligned} \bigl(\lambda\rho_{1}+(1-\lambda)\rho_{2} \bigr) (t) = \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s) \bigl(\lambda f_{1}(s)+(1-\lambda) f_{2}(s) \bigr)\,ds. \end{aligned}$$
Since \(N_{F,x}\) is convex, we have \(\lambda\rho_{1}+(1-\lambda)\rho_{2}\in {\mathcal {B}}x\).
Step 3. \({\mathcal {B}}\) sends bounded sets to bounded sets in \(C([-\tau,b],X)\).
It suffices to prove that there exists a constant \(l>0\) such that for each \(\rho\in{\mathcal {B}}x\), \(x\in B_{k_{0}}=\{x\in C([-\tau,b],X),\|x\|_{[-\tau,b]}\leq k_{0}\}\), we have \(\|\rho\|_{[-\tau,b]}\leq l\).
Let \(\rho\in{\mathcal {B}}x\). Then there exists \(f\in N_{F,x}\) such that for \(t\in[0,b]\), we have
$$\begin{aligned} \rho(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds. \end{aligned}$$
Then, for \(t\in[0,b]\), we have
$$ \bigl|\rho(t)\bigr|= \biggl| \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds \biggr| \leq\frac{q M}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}\bigl|f(s)\bigr|\,ds. $$
(4.2)
From (H2), for \(t\in[0,b]\), we have
$$\begin{aligned} \int_{0}^{t}(t-s)^{q-1}\bigl|f(s)\bigr|\,ds \leq& \biggl( \int_{0}^{t}(t-s)^{\frac{q-1}{1-q_{1}}}\,ds \biggr)^{1-q_{1}}\|m\|_{L^{\frac {1}{q_{1}}}[0,t]}+\overline{r}k_{0} \int_{0}^{t}(t-s)^{q-1}\,ds \\ \leq& \frac{M_{1}}{(1+a)^{1-q_{1}}}b^{(1+a)(1-q_{1})}+\frac{\overline {r}k_{0}b^{q}}{q}, \end{aligned}$$
(4.3)
where \(a=\frac{q-1}{1-q_{1}}\in(-1,0)\), \(M_{1}=\|m\|_{L^{\frac {1}{q_{1}}}[0,b]}\), and \(\overline{r}=\sup\{r(t),t\in[0,b]\}\).
Then from (4.2) and (4.3) we get that
$$\begin{aligned} \|\rho\|_{[-\tau,b]} \leq& \|\rho\|_{[-\tau,0]}+\|\rho\|_{[0,b]} \\ \leq&0+\frac{q M}{\Gamma(1+q)} \biggl[\frac {M_{1}}{(1+a)^{1-q_{1}}}b^{(1+a)(1-q_{1})}+ \frac{\overline{r}k_{0}b^{q}}{q} \biggr] \\ :=& l. \end{aligned}$$
Step 4. \({\mathcal {B}}\) maps bounded sets to equicontinuous sets of \(C([-\tau,b],X)\).
Let \(t_{1}, t_{2}\in[0,b]\), \(t_{1}< t_{2}\), and let \(B_{k_{0}}=\{x\in C([-\tau ,b],X),\|x\|_{[-\tau,b]}\leq k_{0}\}\) be a bounded set of \(C([-\tau,b],X)\). For each \(x\in B_{k_{0}}\) and \(\rho\in {\mathcal {B}}x\), there exists \(f\in N_{F,x}\) such that
$$\begin{aligned} \rho(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds. \end{aligned}$$
Then,
$$\begin{aligned} \bigl|\rho(t_{2})-\rho(t_{1})\bigr| =& \biggl| \int_{0}^{t_{2}}(t_{2}-s)^{q-1} \frak{T}_{q}(t_{2}-s)f(s)\,ds - \int_{0}^{t_{1}}(t_{1}-s)^{q-1} \frak{T}_{q}(t_{1}-s)f(s)\,ds \biggr| \\ \leq& \biggl| \int_{t_{1}}^{t_{2}}(t_{2}-s)^{q-1} \frak{T}_{q}(t_{2}-s)f(s)\,ds \biggr| \\ &{}+ \biggl| \int_{0}^{t_{1}} \bigl[(t_{2}-s)^{q-1}-(t_{1}-s)^{q-1} \bigr]\frak {T}_{q}(t_{2}-s)f(s)\,ds \biggr| \\ &{}+ \biggl| \int_{0}^{t_{1}}(t_{1}-s)^{q-1} \bigl[\frak{T}_{q}(t_{2}-s)-\frak {T}_{q}(t_{1}-s) \bigr]f(s)\,ds \biggr| \\ =& I_{1}+I_{2}+I_{3}, \end{aligned}$$
where
$$\begin{aligned}& I_{1}= \biggl| \int_{t_{1}}^{t_{2}}(t_{2}-s)^{q-1} \frak {T}_{q}(t_{2}-s)f(s)\,ds \biggr|, \\& I_{2}= \biggl| \int_{0}^{t_{1}} \bigl[(t_{2}-s)^{q-1}-(t_{1}-s)^{q-1} \bigr]\frak {T}_{q}(t_{2}-s)f(s)\,ds \biggr|, \\& I_{3} = \biggl| \int_{0}^{t_{1}}(t_{1}-s)^{q-1} \bigl[\frak{T}_{q}(t_{2}-s)-\frak {T}_{q}(t_{1}-s) \bigr]f(s)\,ds \biggr|. \end{aligned}$$
By using a similar argument as that used in (4.3), we can conclude that
$$\begin{aligned}& I_{1}\leq \frac{q M}{\Gamma (1+q)} \biggl(\frac{M_{1}}{(1+a)^{1-q_{1}}}(t_{2}-t_{1})^{(1+a)(1-q_{1})}+ \frac {\overline{r}k_{0}(t_{2}-t_{1})^{q}}{q} \biggr), \\& \begin{aligned}[b] I_{2}\leq{}& \frac{q M}{\Gamma(1+q)} \biggl[ \biggl( \int _{0}^{t_{1}} \bigl((t_{1}-s)^{q-1}-(t_{2}-s)^{q-1} \bigr)^{\frac{1}{1-q_{1}}}\,ds \biggr)^{1-q_{1}}\|m\|_{L^{\frac{1}{q_{1}}}[0,t_{1}]} \\ &{}+\overline{r}k_{0} \int_{0}^{t_{1}}(t_{1}-s)^{q-1}-(t_{2}-s)^{q-1} \,ds \biggr] \\ \leq{}&\frac{qM}{\Gamma (1+q)} \biggl[ M_{1} \biggl( \int _{0}^{t_{1}} \bigl((t_{1}-s)^{a}-(t_{2}-s)^{a} \bigr)\,ds \biggr)^{1-q_{1}} \\ &{} +\overline{r}k_{0} \biggl(\frac{(t_{2}-t_{1})^{q}}{q}-\frac{t_{2}^{q}}{q}+ \frac {t_{1}^{q}}{q} \biggr) \biggr] \\ ={}&\frac{qM}{\Gamma (1+q)} \biggl(\frac {M_{1}}{(1+a)^{1-q_{1}}} \bigl(t_{1}^{1+a}-t_{2}^{1+a}+(t_{2}-t_{1})^{1+a} \bigr)^{1-q_{1}} +\frac{\overline{r}k_{0}}{q} \bigl((t_{2}-t_{1})^{q}-t_{2}^{q}+t_{1}^{q} \bigr) \biggr) \\ \leq{}&\frac{MM_{0}}{\Gamma(1+q)} \biggl( \frac{M_{1}}{(1+a)^{1-q_{1}}}(t_{2}-t_{1})^{(1+a)(1-q_{1})}+ \frac{\overline {r}k_{0}}{q}(t_{2}-t_{1})^{q} \biggr). \end{aligned} \end{aligned}$$
Hence, \(\lim_{t_{2}\rightarrow t_{1}}I_{1}=0\) and \(\lim_{t_{2}\rightarrow t_{1}}I_{2}=0\) independently of \(x\in B_{k_{0}}\).
On the other hand,
$$\begin{aligned} I_{3} \leq \int_{0}^{t_{1}}(t_{1}-s)^{q-1}\bigl\| \frak{T}_{q}(t_{2}-s)-\frak{T}_{q}(t_{1}-s) \bigr\| \bigl|f(s)\bigr|\,ds. \end{aligned}$$
Hypothesis (H1) and Lemma 4.3 imply that \(\frak{T}_{q} (t)\) is continuous for \(t>0\). Then we get that \(I_{3}\) tends to 0 independently of \(x\in B_{k_{0}}\) as \(t_{2}\rightarrow t_{1}\).
Consequently, \(|\rho(t_{2})-\rho(t_{1})|\rightarrow0\) independently of \(x\in B_{k_{0}}\) as \(t_{2}\rightarrow t_{1}\), which means that \({\mathcal {B}} (B_{k_{0}})\) is equicontinuous.
Step 5. For each \(t\in[0,b]\), \(V(t)=\{({\mathcal {B}}x)(t), x\in B_{k_{0}}\}\) is relatively compact in X.
Obviously, \(V(0)=\{0\}\) is relatively compact in X. Let \(0< t\leq b\) be fixed. For \(x\in B_{k_{0}}\) and \(\rho\in{\mathcal {B}} x\), there exists \(f\in N_{F,x}\) such that
$$\begin{aligned} \rho(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds. \end{aligned}$$
For arbitrary \(\epsilon\in(0,t)\) and \(\delta>0\), define the operator \(\Pi_{\epsilon,\delta}\) on \(B_{k_{0}}\) by
$$\begin{aligned} (\Pi_{\epsilon,\delta}x) (t) =& q \int_{0}^{t-\epsilon} \int_{\delta }^{\infty}\theta(t-s)^{q-1} \xi_{q}(\theta)T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \\ =&q \int_{0}^{t-\epsilon} \int_{\delta}^{\infty}\theta(t-s)^{q-1}\xi _{q}(\theta)T \bigl(\epsilon^{q}\delta \bigr)T \bigl((t-s)^{q}\theta-\epsilon^{q}\delta \bigr) f(s)\,d \theta \,ds \\ =&T \bigl(\epsilon^{q}\delta \bigr)q \int_{0}^{t-\epsilon} \int_{\delta}^{\infty }\theta(t-s)^{q-1} \xi_{q}(\theta)T \bigl((t-s)^{q}\theta-\epsilon^{q} \delta \bigr)f(s)\,d\theta \,ds. \end{aligned}$$
According to the compactness of \(T(t)\), \(t>0\), we get that the set \(V_{\epsilon,\delta}(t)=\{(\Pi_{\epsilon,\delta}x)(t),x\in B_{k_{0}} \}\) is relatively compact in X for all \(\epsilon\in(0,t)\) and \(\delta>0\). Moreover, for every \(x\in B_{k_{0}}\), we have
$$\begin{aligned} &\bigl|({\mathcal {B}} x) (t)-(\Pi_{\epsilon,\delta}x) (t)\bigr| \\ &\quad= \biggl| q \int_{0}^{t} \int_{0}^{\delta}\theta(t-s)^{q-1} \xi_{q}(\theta )T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \\ &\qquad{}+q \int_{0}^{t} \int_{\delta}^{\infty}\theta(t-s)^{q-1} \xi_{q}(\theta )T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \\ &\qquad{}-q \int_{0}^{t-\epsilon} \int_{\delta}^{\infty}\theta(t-s)^{q-1}\xi _{q}(\theta)T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \biggr| \\ &\quad\leq q \biggl| \int_{0}^{t} \int_{0}^{\delta}\theta(t-s)^{q-1} \xi_{q}(\theta )T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \biggr| \\ &\qquad{}+q \biggl| \int_{t-\epsilon}^{t} \int_{\delta}^{\infty}\theta (t-s)^{q-1} \xi_{q}(\theta)T \bigl((t-s)^{q}\theta \bigr) f(s)\,d\theta \,ds \biggr| \\ &\quad\leq qM \int_{0}^{t}(t-s)^{q-1}\bigl|f(s)\bigr|\,ds \int_{0}^{\delta}\theta \xi_{q}(\theta)\,d \theta\\ &\qquad{}+qM \int_{t-\epsilon}^{t}(t-s)^{q-1}\bigl|f(s)\bigr|\,ds \int_{0}^{\infty}\theta \xi_{q}(\theta)\,d \theta. \end{aligned}$$
In view of (4.3), we have
$$\begin{aligned} &\bigl|({\mathcal {B}} x) (t)-(\Pi_{\epsilon,\delta}x) (t)\bigr| \\ &\quad\leq qM \int_{0}^{\delta}\theta \xi_{q}(\theta)\,d \theta \biggl(\frac {M_{1}}{(1+a)^{1-q_{1}}}b^{(1+a)(1-q_{1})}+\frac{\overline{r}k_{0}b^{q}}{q} \biggr) \\ &\qquad{}+\frac{qM}{\Gamma(1+q)} \biggl[ \biggl( \int_{t-\epsilon}^{t} (t-s)^{\frac{q-1}{1-q_{1}}}\,ds \biggr)^{1-q_{1}}\|m\|_{L^{\frac {1}{q_{1}}}[t-\epsilon,t]}+\overline{r}k_{0} \int_{t-\epsilon}^{t} (t-s)^{q-1}\,ds \biggr] \\ &\quad\leq qM \int_{0}^{\delta}\theta \xi_{q}(\theta)\,d \theta \biggl(\frac {M_{1}}{(1+a)^{1-q_{1}}}b^{(1+a)(1-q_{1})}+\frac{\overline{r}k_{0}b^{q}}{q} \biggr) \\ &\qquad{}+\frac{qM}{\Gamma(1+q)} \biggl(\frac{M_{1}}{(1+a)^{1-q_{1}}}\epsilon ^{(1+a)(1-q_{1})}+ \frac{\overline{r}k_{0}\epsilon^{q}}{q} \biggr). \end{aligned}$$
From \(\lim_{\delta\rightarrow0}\int_{0}^{\delta}\theta\xi_{q}(\theta )\,d\theta=0\) we conclude that there exist relatively compact sets arbitrarily approximating the sets \(V(t)\), \(t>0\). Hence, the sets \(V(t)\) are relatively compact in X for all \(t>0\).
Step 6. \({\mathcal {B}}\) has a closed graph.
Let \(x^{n}\rightarrow x^{\ast}\), \(\rho_{n}\in{\mathcal {B}}x^{n}\), and \(\rho_{n}\rightarrow\rho_{\ast}\) as \(n\rightarrow\infty\). We shall prove that \(\rho_{\ast}\in {\mathcal {B}} x^{\ast}\). Since \(\rho_{n}\in{\mathcal {B}} x^{n}\), there exists \(f_{n}\in N_{F,x^{n}}\) such that
$$\rho_{n}(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f_{n}(s)\,ds,\quad t\in [0,b]. $$
We need to prove that there exists \(f_{\ast}\in N_{F,x^{\ast}}\) such that
$$\rho_{\ast}(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f_{\ast}(s)\,ds,\quad t\in[0,b]. $$
Consider the continuous operator \({\mathcal {T}}: L^{1}([0,b],X)\rightarrow C([0,b],X)\) defined by
$$( {\mathcal {T}}f) (t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds. $$
We can easily obtain that \(\mathcal{T}\) is continuous. On the other hand,
$$\begin{aligned} \bigl|\rho_{n}(t)-\rho_{\ast}(t)\bigr|\leq\|\rho_{n}- \rho_{\ast}\|\rightarrow0\quad \mbox{as } n\rightarrow\infty. \end{aligned}$$
From Theorem 2.6 it follows that \({\mathcal {T}}\circ N_{F}\) is a closed graph operator. Moreover, we have that
$$\rho_{n}\in {\mathcal {T}}(N_{F,x^{n}}). $$
Since \(x_{n}\rightarrow x^{\ast}\), by Theorem 2.6 there exists \(f_{\ast}\in N_{F,x^{\ast}}\) such that
$$\rho_{\ast}(t)= \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f_{\ast}(s)\,ds, \quad t\in[0,b]. $$
This implies that \(\rho_{\ast}\in{\mathcal {B}} x^{\ast}\).
Therefore, the multivalued map \({\mathcal {B}}\) is completely continuous and u.s.c. with convex closed values.
Step 7. The set \(U=\{u\in C([-\tau,0],X)|u\in\lambda{ \mathcal {A}}u+\lambda{\mathcal {B}}u, 0<\lambda<1\}\) is bounded.
Let \(x\in U\), then \(x\in\lambda\Phi x\) for some \(0<\lambda< 1\). Thus, there exists \(f\in N_{F,x}\) such that for \(t\in[0,b]\),
$$\begin{aligned} x(t) =&\lambda \biggl(\frak{S}_{q}(t) \bigl[\phi(0)-g(0,\phi) \bigr]+g(t,x_{t}) + \int_{0}^{t}(t-s)^{q-1}A \frak{T}_{q}(t-s)g(s,x_{s})\,ds \\ &{}+ \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds \biggr). \end{aligned}$$
By (H2) and (H3), for each \(t\in[0,b]\), we have
$$\begin{aligned} \bigl|x(t)\bigr| =&\lambda \biggl|\frak{S}_{q}(t) \bigl[\phi(0)-g(0,\phi) \bigr]+g(t,x_{t}) + \int_{0}^{t}(t-s)^{q-1}A \frak{T}_{q}(t-s)g(s,x_{s})\,ds \\ &{}+ \int_{0}^{t}(t-s)^{q-1} \frak{T}_{q}(t-s)f(s)\,ds \biggr| \\ \leq& \bigl\| \frak{S}_{q}(t)\bigr\| \bigl|\phi(0)-g(0,\phi)\bigr|+\bigl\| A^{-\beta} \bigr\| \bigl|A^{\beta}g(t,x_{t})\bigr| \\ &{}+ \int_{0}^{t}(t-s)^{q-1}\bigl\| A^{1-\beta} \frak{T}_{q}(t-s)\bigr\| \bigl|A^{\beta}g(s,x_{s})\bigr|\,ds\\ &{}+ \int_{0}^{t}(t-s)^{q-1}\bigl\| \frak{T}_{q}(t-s)\bigr\| \bigl|f(s)\bigr|\,ds \\ \leq& M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr) +\bigl\| A^{-\beta}\bigr\| \bigl(L_{1} \|x_{t}\|_{[-\tau,0]}+L_{2}\bigr) \\ &{}+ \int_{0}^{t}(t-s)^{q-1} \frac{qC_{1-\beta}\Gamma(1+\beta )}{(t-s)^{q(1-\beta)}\Gamma(1+q\beta)}\bigl(L_{1}\|x_{s}\|_{[-\tau,0]}+L_{2}\bigr) \,ds \\ &{}+\frac{qM}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}\bigl|f(s)\bigr|\,ds \\ \leq&M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr) +\bigl\| A^{-\beta}\bigr\| \bigl(L_{1} \|x_{t}\|_{[-\tau ,0]}+L_{2}\bigr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t}(t-s)^{q\beta-1}\|x_{s} \|_{[-\tau,0]}\,ds \\ &{}+\frac{qL_{2}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t}(t-s)^{q\beta-1}\,ds \\ &{}+\frac{qM}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}m(s)\,ds+ \frac{qM}{\Gamma (1+q)} \int_{0}^{t}(t-s)^{q-1}r(s) \|x_{s}\|_{[-\tau,0]}\,ds \\ \leq&M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr) +\bigl\| A^{-\beta}\bigr\| \bigl(L_{1} \|x_{t}\|_{[-\tau ,0]}+L_{2}\bigr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t}(t-s)^{q\beta-1}\|x_{s} \|_{[-\tau,0]}\,ds +\frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta )} \\ &{}+\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})}+\frac {qM\overline{r}}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}\|x_{s} \|_{[-\tau,0]}\,ds. \end{aligned}$$
Consider the function
$$\mu(t)=\max \bigl\{ \bigl|x(s)\bigr|:s\in[-\tau,t] \bigr\} ,\quad t\in[0,b]. $$
Let \(t^{\ast}\in[-\tau,t]\) be such that \(\mu(t)=|x(t^{\ast})|\). If \(t^{\ast}\in[0,t]\), then for \(t\in[0,b]\), we have
$$\begin{aligned} \mu(t) \leq& M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr) +\bigl\| A^{-\beta}\bigr\| \bigl(L_{1}\mu(t)+L_{2} \bigr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t^{\ast}} \bigl(t^{\ast}-s \bigr)^{q\beta-1}\mu(s)\,ds +\frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta )} \\ &{}+\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})}+\frac {qM\overline{r}}{\Gamma(1+q)} \int_{0}^{t^{\ast}} \bigl(t^{\ast}-s \bigr)^{q-1}\mu(s)\,ds. \end{aligned}$$
Since \(\mu(t)\) is nondecreasing, for \(t^{\ast}\in[0,t]\), we have
$$\begin{aligned} \int_{0}^{t^{\ast}} \bigl(t^{\ast}-s \bigr)^{q-1}\mu(s)\,ds&= \int_{0}^{t^{\ast}}s^{q-1}\mu \bigl(t^{\ast}-s \bigr)\,ds\\ &\leq \int_{0}^{t}s^{q-1}\mu(t-s)\,ds= \int_{0}^{t}(t-s)^{q-1}\mu(s)\,ds. \end{aligned}$$
Therefore,
$$\begin{aligned} \mu(t) \leq& M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr) +\bigl\| A^{-\beta}\bigr\| \bigl(L_{1}\mu(t)+L_{2} \bigr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t}(t-s)^{q\beta-1}\mu(s)\,ds + \frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta )} \\ &{}+\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})}+\frac {qM\overline{r}}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}\mu(s)\,ds. \end{aligned}$$
If \(t^{\ast}\in[-\tau,0]\), then \(\mu(t)=\|\phi\|_{[-\tau,0]}\).
Therefore,
$$\begin{aligned} \mu(t) \leq&\|\phi\|_{[-\tau,0]}+ M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr)+ \bigl\| A^{-\beta }\bigr\| \bigl(L_{1}\mu(t)+L_{2} \bigr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{\Gamma(1+q\beta)} \int _{0}^{t}(t-s)^{q\beta-1}\mu(s)\,ds + \frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta)} \\ &{}+\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})}+\frac {qM\overline{r}}{\Gamma(1+q)} \int_{0}^{t}(t-s)^{q-1}\mu(s)\,ds. \end{aligned}$$
From \(\|A^{-\beta}\|L_{1}<1\) we get that
$$\begin{aligned} \mu(t) \leq&\frac{1}{1-\|A^{-\beta}\|L_{1}} \biggl(\|\phi\|_{[-\tau,0]}+ M \bigl(\bigl| \phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr)+L_{2}\bigl\| A^{-\beta}\bigr\| \\ &{}+\frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta)} +\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})} \biggr) \\ &{}+\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{(1-\|A^{-\beta}\|L_{1})\Gamma (1+q\beta)} \int_{0}^{t}(t-s)^{q\beta-1}\mu(s)\,ds \\ &{}+\frac{qM\overline{r}}{(1-\|A^{-\beta}\|L_{1})\Gamma(1+q)} \int _{0}^{t}(t-s)^{q-1}\mu(s)\,ds \\ =& \overline{C}+\overline{G} \int_{0}^{t}(t-s)^{q\beta-1}u(s)\,ds+\overline {K} \int_{0}^{t}(t-s)^{q-1}u(s)\,ds, \end{aligned}$$
where
$$\begin{aligned}& \begin{aligned}[b] \overline{C}={}&\frac{1}{1-\|A^{-\beta}\|L_{1}} \biggl(\|\phi \|_{[-\tau ,0]}+ M \bigl(\bigl|\phi(0)\bigr|+\bigl|g(0,\phi)\bigr| \bigr)+L_{2} \bigl\| A^{-\beta}\bigr\| \\ &{}+\frac{L_{2}C_{1-\beta}b^{q\beta}\Gamma(1+\beta)}{\beta\Gamma(1+q\beta)} +\frac{qMM_{1}}{(1+a)^{1-q_{1}}\Gamma(1+q)}b^{(1+a)(1-q_{1})} \biggr), \end{aligned} \\& \overline{G}=\frac{qL_{1}C_{1-\beta}\Gamma(1+\beta)}{(1-\|A^{-\beta}\| L_{1})\Gamma(1+q\beta)}, \qquad \overline{K}=\frac{qM\overline{r}}{(1-\|A^{-\beta}\|L_{1})\Gamma(1+q)}. \end{aligned}$$
Then by Theorem 3.2 we have
$$\begin{aligned} \mu(t) \leq& \overline{C}+ \int_{0}^{t}\sum_{n=1}^{\infty} \sum_{k=0}^{n}C_{n}^{k} \frac{(\overline{G}\Gamma(q\beta))^{k}(\overline{K}\Gamma (q))^{n-k}}{\Gamma(kq\beta+(n-k)q)} (t-s)^{kq\beta+(n-k)q-1}\overline{C}\,ds \\ \leq& \overline{C}+\overline{C}\sum_{n=1}^{\infty} \sum_{k=0}^{n} C_{n}^{k} \frac{(\overline{G}\Gamma(q\beta))^{k}(\overline{K}\Gamma (q))^{n-k}b^{kq\beta+(n-k)q}}{(kq\beta+(n-k)q)\Gamma(kq\beta+(n-k)q)}. \end{aligned}$$
Therefore, we obtain that
$$ \|x\|_{[-\tau,b]}=\mu(b)\leq\overline{C}+\overline{C}\sum _{n=1}^{\infty }\sum_{k=0}^{n}C_{n}^{k} \frac{(\overline{G}\Gamma(q\beta))^{k}(\overline{K}\Gamma (q))^{n-k}b^{kq\beta+(n-k)q}}{ (kq\beta+(n-k)q)\Gamma(kq\beta+(n-k)q)}. $$
This shows that U is bounded.
As a result of Theorem 2.7, we obtain that Φ has a fixed point, which is the mild solution of system (1.1). This completes the proof. □