Let us recall some basic definitions on multivalued maps [33, 34].
For a normed space \((X, \|\cdot\|)\), let \(P_{\mathrm{cl}}(X)=\{Y \in{ F}(X) : Y \mbox{ is closed}\}\), \(P_{\mathrm{b}}(X)=\{Y \in{ F}(X) : Y \mbox{ is bounded}\}\), \(P_{\mathrm{cp}}(X)=\{Y \in{ F}(X) : Y \mbox{ is compact}\}\), and \(P_{\mathrm{cp}, \mathrm{c}}(X) =\{Y \in{ F}(X) : Y \mbox{ is compact and convex}\}\). A multivalued map \(G : X \to{ F}(X)\) is convex (closed)-valued if \(G(x)\) is convex (closed) for all \(x \in X\). A map G is bounded on bounded sets if \(G(\mathbb{B}) = \bigcup_{x \in\mathbb{B}}G(x)\) is bounded in X for all \(\mathbb{B} \in P_{\mathrm{b}}(X)\) (i.e., \(\sup_{x \in \mathbb{B}}\{\sup\{|y| : y \in G(x)\}\} < \infty\)). A map G is called upper semicontinuous (u.s.c.) on X if for each \(x_{0} \in X\), the set \(G(x_{0})\) is a nonempty closed subset of X and if for each open set N of X containing \(G(x_{0})\), there exists an open neighborhood \(\mathcal{N}_{0}\) of \(x_{0}\) such that \(G(\mathcal{N}_{0}) \subseteq N\); G is said to be completely continuous if \(G(\mathbb{B})\) is relatively compact for every \(\mathbb{B} \in P_{\mathrm{b}}(X)\). If a multivalued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph, that is, \(x_{n} \to x_{*}\), \(y_{n} \to y_{*}\), \(y_{n} \in G(x_{n})\) imply \(y_{*} \in G(x_{*})\). A map G has a fixed point if there is \(x \in X\) such that \(x \in G(x)\). The fixed point set of a multivalued operator G will be denoted by FixG. A multivalued map \(G : [0;1] \to{\mathcal{P}}_{\mathrm{cl}}(\mathbb{R})\) is said to be measurable if for every \(y \in \mathbb{R}\), the function \(t \mapsto d(y,G(t)) = \inf\{|y-z|: z \in G(t)\}\) is measurable.
Definition 4.1
A function \(x \in \mathit{AC}^{n}([0,1], \mathbb{R})\) satisfying the conditions \(x(0)+x'(0)= h(x)\), \(\int _{0}^{\eta}x(t)\,dt=\xi\), \(x''(0)=x'''(0)=\cdots=x^{(n-1)}(0)=0\) is said to be a solution of problem (1.2) if there exists a function \(f \in L^{1}([0,1],\mathbb{R})\) such that \(f(t) \in F(t, x(t), {}^{\mathrm{c}}D^{\beta}x(t))\) a.e. on \([0,1]\) and
$$\begin{aligned} x(t) =& \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}f(s)\,ds+ \frac{2(1-t)}{\eta(2-\eta)}\xi+\frac{2t-\eta }{2-\eta}h(x) \\ &{}+\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac {(\eta-s)^{\alpha}}{\Gamma(\alpha+1)}f(s)\,ds. \end{aligned}$$
(4.1)
Definition 4.2
A multivalued map \(F : [0,1] \times\mathbb{R}\times\mathbb{R} \to{\mathcal{P}}(\mathbb{R})\) is said to be a Carathéodory function if
-
(i)
\(t \mapsto F(t,x,y)\) is measurable for each \(x ,y\in\mathbb{R}\);
-
(ii)
\((x, y) \mapsto F(t,x,y)\) is upper semicontinuous for almost all \(t\in[0,1]\).
Further, a Carathéodory function F is called \(L^{1}\)-Carathéodory if
-
(iii)
for each \(a > 0\), there exists \(\varphi_{a} \in L^{1}([0,1],\mathbb{R}^{+})\) such that
$$\bigl\Vert F (t, x, y)\bigr\Vert = \sup\bigl\{ |v| : v \in F (t, x, y)\bigr\} \le\varphi_{a} (t) $$
for all \(\|x\|, \|y\| \le a\) and for a.e. \(t \in[0,1]\).
For each \(y \in C([0,1], \mathbb{R})\), define the set of selections of F by
$$S_{F,y} := \bigl\{ v \in L^{1}\bigl([0,1],\mathbb{R}\bigr) : v (t) \in F \bigl(t, y(t), {}^{\mathrm{c}}D^{\beta}y(t)\bigr) \mbox{ for a.e. } t \in [0,1]\bigr\} . $$
The following lemma will be used in the sequel.
Lemma 4.1
[35]
Let
X
be a Banach space. Let
\(F : [0, 1] \times X\times X \to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(X)\)
be an
\(L^{1}\)-Carathéodory multivalued map, and let Θ be a linear continuous mapping from
\(L^{1}([0,1],X)\)
to
\(C([0,1],X)\). Then the operator
$$\Theta\circ S_{F} : C\bigl([0,1],X\bigr) \to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}} \bigl(C\bigl([0,1],X\bigr)\bigr), \qquad x \mapsto(\Theta\circ S_{F}) (x) = \Theta( S_{F,x}), $$
is a closed graph operator in
\(C([0,1],X) \times C([0,1],X)\).
To prove our main result in this section, we use the following form of the nonlinear alternative for contractive maps ([36], Corollary 3.8).
Theorem 4.1
Let
X
be a Banach space, and
\(\mathcal{D}\)
a bounded neighborhood of
\(0\in X\). Let
\(Z_{1}:X\to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(X)\)
and
\(Z_{2}:\bar{\mathcal{D}}\to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(X)\)
be two multivalued operators satisfying
-
(a)
\(Z_{1}\)
is contraction, and
-
(b)
\(Z_{2}\)
is u.s.c. and compact.
Then, if
\(G=Z_{1}+Z_{2}\), then either
-
(i)
G
has a fixed point in
\(\bar{\mathcal{D}}\), or
-
(ii)
there are a point
\(u\in\partial\mathcal{D}\)
and
\(\lambda\in(0,1)\)
with
\(u\in\lambda G(u)\).
Theorem 4.2
Assume that (A1) and (A2) hold. In addition, we suppose that:
- (H1):
-
\(F : [0,1] \times\mathbb{R} \times\mathbb{R} \to {\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(\mathbb{R})\)
is an
\(L^{1}\)-Carathéodory multivalued map;
- (H2):
-
there exist continuous nondecreasing functions
\(\psi_{i} : [0,\infty) \to(0,\infty)\), \(i=1,2\), and a function
\(p \in C([0,1],\mathbb{R}^{+})\)
such that
$$\bigl\Vert F(t,x,y)\bigr\Vert _{\mathcal{P}}:=\sup\bigl\{ |v|: v \in F(t,x, y)\bigr\} \le p(t)\bigl[\psi _{1}\bigl(\vert x\vert \bigr)+ \psi_{2}\bigl(\vert y\vert \bigr)\bigr] $$
for each
\((t,x,y) \in[0,1] \times\mathbb{R}\times\mathbb{R}\);
- (H3):
-
there exists a number
\(M>0\)
such that
$$ \frac{(1- \ell \Lambda_{3})M}{ \Lambda_{1} \|p\| [\psi_{1}(M)+ \psi_{2}(M)]+\Lambda_{2}} > 1, \quad \ell\Lambda_{3}< 1, $$
(4.2)
where
\(\Lambda_{i}\), \(i=1,2,3\), are defined in (3.4) and (3.5).
Then the boundary value problem (1.1) has at least one solution on
\([0,1]\).
Proof
To transform problem (1.1) into a fixed point problem, we introduce the operator \({\mathcal{N}}:X\rightarrow{\mathcal{P}}(X)\) as follows:
$${\mathcal{N}}(x)= \left\{ \begin{aligned} &h \in X: \\ &\quad h(t) = \left\{ \begin{aligned} &\int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}f(s)\,ds+ \frac{2(1-t)}{\eta(2-\eta)}\xi+\frac{2t-\eta}{2-\eta }h(x) \\ &\quad {}+\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s)\,ds, \quad f \in S_{F,x} \end{aligned} \right . \end{aligned} \right\}. $$
Now, we define the operator \({\mathcal{A}}_{1}: X\rightarrow X\) by
$$ {\mathcal{A}}_{1} x(t) = \frac{2(1-t)}{\eta(2-\eta)}\xi + \frac{2t-\eta}{2-\eta}h(x) $$
(4.3)
and the multivalued operator \({\mathcal{A}}_{2}: X\rightarrow {\mathcal{P}}(X)\) by
$$ {\mathcal{A}}_{2}(x)= \left\{ \begin{aligned} &h \in C\bigl([0,1], \mathbb{R}\bigr): \\ &\quad h(t) = \left\{ \begin{aligned} &\int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s)\,ds \end{aligned} \right . \end{aligned} \right\}. $$
(4.4)
Observe that \({\mathcal{N}}={\mathcal{A}}_{1}+{\mathcal{A}}_{2}\). We shall show that the operators \({\mathcal{A}}_{1}\) and \({\mathcal{A}}_{2}\) satisfy all the conditions of Theorem 4.1 on \([0,1]\). Also, we establish that the operators \({\mathcal{A}}_{1}\) and \({\mathcal{A}}_{2}\) are such that \({\mathcal{A}}_{1},{\mathcal{A}}_{2}: B_{r}\to{\mathcal{P}}_{\mathrm{cp},\mathrm{c}}(X)\), where \(B_{r} = \{x \in X: \|x\|_{X} \le r \}\) is a bounded set in \(C([0,1], \mathbb{R})\). First, we prove that \({\mathcal{A}}_{2}\) is compact-valued on \(B_{r}\). Note that the operator \({\mathcal{A}}_{2}\) is equivalent to the composition \({\mathcal{L}} \circ S_{F}\), where \({\mathcal{L}} \) is the continuous linear operator from \(L^{1}([0,1], \mathbb{R})\) into X defined by
$${\mathcal{L}} (v) (t) = \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}v(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}v(s)\,ds. $$
Suppose that \(x\in B_{r}\) is arbitrary and let \(\{v_{n}\}\) be a sequence in \(S_{F,x}\). Then, by the definition of \(S_{F,x}\), we have \(v_{n}(t)\in F(t, x(t), {}^{\mathrm{c}}D^{\beta}x(t))\) for almost all \(t\in[0,1]\). Since \(F(t, x(t), {}^{\mathrm{c}}D^{\beta}x(t))\) is compact for all \(t\in J\), there is a convergent subsequence of \(\{v_{n}(t)\}\) (we denote it by \(\{v_{n}(t)\}\) again) that converges in measure to some \(v(t)\in S_{F,x}\) for almost all \(t\in J\). On the other hand, \({\mathcal{L}} \) is continuous, so \({\mathcal{L}} (v_{n})(t)\to{\mathcal{L}} (v)(t)\) pointwise on \([0,1]\).
In order to show that the convergence is uniform, we have to show that \(\{{\mathcal{L}} (v_{n})\}\) is an equicontinuous sequence. Let \(t_{1}, t_{2} \in[0,1]\) with \(t_{1}< t_{2} \). Then, we have
$$\begin{aligned} \bigl\vert {\mathcal{L}} (v_{n}) (t_{2}) - {\mathcal{L}} (v_{n}) (t_{1})\bigr\vert \leq& \biggl\vert \frac{1}{\Gamma(\alpha)} \int_{0}^{t_{1}}\bigl[(t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha-1} \bigr]v_{n}(s)\,ds \\ &{} + \frac{1}{\Gamma(\alpha)} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha -1}v_{n}(s) \,ds \biggr\vert \\ &{}+\frac{2(t_{2}-t_{1})}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta -s)^{\alpha}}{\Gamma(\alpha+1)}\bigl\vert v_{n}(s)\bigr\vert \,ds \\ \le& \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha+1)}\bigl(t_{2}^{\alpha}-t_{1}^{\alpha}\bigr) +\frac{2(t_{2}-t_{1})}{\eta\Gamma(\alpha+2)} \biggr\} . \end{aligned}$$
Continuing this process, we have
$$\begin{aligned} \bigl\vert \bigl({\mathcal{L}}' (v_{n}) (t_{2})\bigr)- \bigl({\mathcal{L}}' (v_{n}) (t_{1}) \bigr)\bigr\vert \leq& \|p\| \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha -\beta+1)} \bigl(t_{2}^{\alpha-\beta-1}-t_{1}^{\alpha-\beta-1} \bigr) \\ &{}+\frac{2 }{\eta\Gamma(\alpha+2)\Gamma(2-\beta)}\bigl(t_{2}^{1-\beta}-t_{1}^{1-\beta } \bigr) \biggr\} . \end{aligned}$$
We observe that the right-hand sides of the above inequalities tend to zero as \(t_{2}\to t_{1}\). Thus, the sequence \(\{{\mathcal{L}} (v_{n})\}\) is equicontinuous, and by using the Arzelá-Ascoli theorem we get that there is a uniformly convergent subsequence. So, there is a subsequence of \(\{v_{n}\}\) (we denote it again by \(\{v_{n}\}\)) such that \({\mathcal{L}} (v_{n})\to{\mathcal{L}} (v)\). Note that \({\mathcal{L}} (v) \in{\mathcal{L}} (S_{F,x})\). Hence, \({\mathcal{A}}_{2}(x) ={\mathcal{L}} (S_{F,x})\) is compact for all \(x\in B_{r}\). So \({\mathcal{A}}_{2}(x)\) is compact.
Now, we show that \({\mathcal{A}}_{2}(x)\) is convex for all \(x\in X\). Let \(z_{1},z_{2}\in{\mathcal{A}}_{2}(x)\). We select \(f_{1},f_{2}\in S_{F,x}\) such that
$$z_{i}(t) = \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f_{i}(s) \,ds +\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f_{i}(s) \,ds, \quad i=1,2 $$
for almost all \(t\in[0,1]\). Let \(0\leq\lambda\leq1\). Then, we have
$$\begin{aligned} \bigl[\lambda z_{1}+(1-\lambda)z_{2}\bigr](t) =& \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma(\alpha)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s)\bigr] \,ds \\ &{}+\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta -s)^{\alpha}}{\Gamma(\alpha+1)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s)\bigr] \,ds. \end{aligned}$$
Since F has convex values, \(S_{F,u}\) is convex, and \(\lambda f_{1}(s) + (1-\lambda)f_{2}(s)\in S_{F,x}\). Thus,
$$\lambda z_{1}+(1-\lambda)z_{2}\in{\mathcal{A}}_{2}(x). $$
Consequently, \({\mathcal{A}}_{2}\) is convex-valued. Obviously, \({\mathcal{A}}_{1}\) is compact and convex-valued.
The rest of the proof consists of several steps and claims.
Step 1. We show that
\({\mathcal{A}}_{1}\)
is a contraction on
\(C([0,1],{\mathbb{R}})\). The proof is similar to that for the operator \({F}_{2}\) in Step 2 of Theorem 3.1.
Step 2. \({\mathcal{A}}_{2}\)
is upper semicontinuous and compact. This will be established in several claims.
Claim I: \({\mathcal{A}}_{2}\)
maps bounded sets into bounded sets in
X. Let \(B_{r} = \{x \in X: \|x\|_{X} \le r \}\) be a bounded set in X. Then, for each \(h \in{\mathcal{A}}_{2} (x)\), \(x \in B_{r}\), there exists \(f \in S_{F,x}\) such that
$$h(t) = \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(q)}f(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s)\,ds. $$
Then, for \(t\in[0,1]\), we have
$$\begin{aligned} \Vert h\Vert \le& \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl\vert f(s)\bigr\vert \,ds +\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}\bigl\vert f(s) \bigr\vert \,ds \\ \le& \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl\vert f(s)\bigr\vert \,ds +\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{1}\frac{(1-s)^{\alpha }}{\Gamma(\alpha+1)}\bigl\vert f(s) \bigr\vert \,ds \\ \le& \Vert p\Vert \bigl[\psi_{1}\bigl(\Vert x\Vert \bigr)+ \psi_{2}\bigl(\bigl\Vert {}^{\mathrm{c}}D^{\beta}x\bigr\Vert \bigr)\bigr] \biggl[\frac{1}{\Gamma(\alpha+1)}+\frac{2}{\eta\Gamma(\alpha+2)} \biggr] \\ \le& \Vert p\Vert \bigl[\psi_{1}\bigl(\Vert x\Vert _{X}\bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X} \bigr)\bigr] \biggl[\frac {1}{\Gamma(\alpha+1)}+\frac{2}{\eta\Gamma(\alpha+2)} \biggr] \\ \le& \Vert p\Vert \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl[\frac{1}{\Gamma(\alpha+1)}+\frac{2}{\eta\Gamma(\alpha +2)} \biggr]. \end{aligned}$$
Considering
$$ h'(t) = \int_{0}^{t}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s) \,ds + \frac{2}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s), $$
we obtain
$$\begin{aligned} {}^{\mathrm{c}}D^{\beta}h(t) =& \int_{0}^{t}\frac{(t-s)^{-\beta}}{\Gamma(1-\beta)} \biggl( \int _{0}^{s}\frac{(s-\tau)^{\alpha-2}}{\Gamma(\alpha-1)}f(\tau) d\tau + \frac{2}{\eta} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha}}{\Gamma (\alpha+1)}f(s) \,ds \biggr) \\ =& \int_{0}^{t}\frac{(t-s)^{\alpha-1-\beta}}{\Gamma(\alpha-\beta)}f(s)\,ds + \frac{2t^{1-\beta}}{\eta\Gamma(2-\beta)} \int_{0}^{\eta}\frac {(\eta-s)^{\alpha}}{\Gamma(\alpha+1)}f(s)\,ds, \end{aligned}$$
and therefore
$$\begin{aligned} \bigl\Vert {}^{\mathrm{c}}D^{\beta}h\bigr\Vert \le& \Vert p \Vert \bigl[\psi_{1}\bigl(\Vert x\Vert _{X}\bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X}\bigr)\bigr] \int_{0}^{t}\frac{(t-s)^{\alpha-1-\beta}}{\Gamma(\alpha-\beta)} \,ds \\ &{}+ \frac{2}{\eta\Gamma(2-\beta)} \int_{0}^{1}\frac{(1-s)^{\alpha }}{\Gamma(\alpha+1)} \,ds \\ \le& \Vert p\Vert \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha-\beta+1)}+\frac{2}{\eta\Gamma (2-\beta)}\frac{1}{\Gamma(\alpha+2)} \biggr\} . \end{aligned}$$
Consequently,
$$\begin{aligned} \Vert {\mathcal{A}}_{2}x\Vert _{X} =&\Vert { \mathcal{A}}_{2}x\Vert +\bigl\Vert {}^{\mathrm{c}}D^{\beta }({ \mathcal{A}}_{2}x)\bigr\Vert \\ \le& \Vert p\Vert \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha +1)}+\frac{2}{\eta\Gamma(\alpha+2)}+\frac{1}{\Gamma(\alpha-\beta +1)} \\ &{}+\frac{2}{\eta\Gamma(2-\beta)}\frac{1}{\Gamma(\alpha+2)} \biggr\} , \end{aligned}$$
and thus the operator \({\mathcal{A}}_{2}(B_{r})\) is uniformly bounded.
Claim II: \({\mathcal{A}}_{2}\)
maps bounded sets into equicontinuous sets. Now let \(0\le t_{1}< t_{2}\le1\). Then we have the following facts:
$$\begin{aligned} \bigl\vert ({\mathcal{A}}_{2}x) (t_{2})-({ \mathcal{A}}_{2}x) (t_{1})\bigr\vert \le&\biggl\vert \frac {1}{\Gamma(\alpha)} \int_{0}^{t_{1}}\bigl[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha -1} \bigr]f(s)\,ds \\ &{}+ \frac{1}{\Gamma(\alpha)} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha -1}f(s) \,ds\biggr\vert \\ &{}+\frac{2}{\eta}(t_{2}-t_{1}) \int_{0}^{1}\frac{(1-s)^{\alpha}}{\Gamma (\alpha+1)}\bigl\vert f(s) \bigr\vert \,ds \\ \le& \|p\| \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha+1)}\bigl(t_{2}^{\alpha}-t_{1}^{\alpha}\bigr) +\frac{2(t_{2}-t_{1})}{\eta\Gamma(\alpha+2)} \biggr\} \end{aligned}$$
and
$$\begin{aligned} \begin{aligned} \bigl\vert {}^{\mathrm{c}}D^{\beta}({\mathcal{A}}_{2}x) (t_{2})- {}^{\mathrm{c}}D^{\beta}({\mathcal{A}}_{2}) (t_{1})\bigr\vert \le{}& \|p\| \bigl[\psi_{1}(r)+ \psi_{2}(r)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha-\beta +1)} \bigl(t_{2}^{\alpha-\beta-1}-t_{1}^{\alpha-\beta-1} \bigr) \\ &{}+\frac{2 }{\eta\Gamma(\alpha+2)\Gamma(2-\beta)}\bigl(t_{2}^{1-\beta}-t_{1}^{1-\beta } \bigr) \biggr\} . \end{aligned} \end{aligned}$$
Hence, we have
$$\bigl\Vert ({\mathcal{A}}_{2} x) (t_{2})-({ \mathcal{A}}_{2} x) (t_{1})\bigr\Vert _{X}\to0 \quad \mbox{as } t_{2}\to t_{1}. $$
Thus, \({\mathcal{A}}_{2}\) is equicontinuous. Therefore, by the Ascoli-Arzelá theorem it follows that \({\mathcal{A}}_{2}: X \to{\mathcal{P}}(X)\) is completely continuous.
Claim III: \({\mathcal{A}}_{2}\)
has a closed graph. Let \(x_{n} \to x_{*}\), \(h_{n} \in {\mathcal{A}}_{2} (x_{n})\), and \(h_{n} \to h_{*}\). Then we need to show that \(h_{*} \in {\mathcal{A}}_{2}(x_{*})\). Associated with \(h_{n} \in{\mathcal{A}}_{2} (x_{n})\), there exists \(f_{n} \in S_{F,x_{n}}\) such that for each \(t \in[0,1]\),
$$h_{n}(t) = \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma(q)}f_{n}(s)\,ds +\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f_{n}(s) \,ds. $$
Thus, it suffices to show that there exists \(f_{*} \in S_{F,x_{*}}\) such that for each \(t \in[0,1]\),
$$h_{*}(t) = \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma(q)}f_{*}(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f_{*}(s)\,ds. $$
Let us consider the linear operator \(\Theta: L^{1}([0,1], \mathbb{R}) \to C([0,1], \mathbb{R})\) given by
$$f \mapsto\Theta(f) (t) = \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(q)}f(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s)\,ds. $$
Observe that
$$\begin{aligned}& \bigl\Vert h_{n}(t)-h_{*}(t)\bigr\Vert \\& \quad =\biggl\Vert \int_{0}^{t}\frac{(t-s)^{q-1}}{\Gamma(q)}\bigl(f_{n}(s)-f_{*}(s) \bigr)\,ds +\frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}\bigl(f_{n}(s)-f_{*}(s) \bigr) \,ds\biggr\Vert \to0 \end{aligned}$$
as \(n\to\infty\).
Thus, it follows by Lemma 4.1 that \(\Theta\circ S_{F}\) is a closed graph operator. Further, we have \(h_{n}(t) \in \Theta(S_{F,x_{n}})\). Since \(x_{n} \to x_{*}\), we have
$$h_{*}(t) = \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma(q)}f_{*}(s)\,ds + \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f_{*}(s)\,ds $$
for some \(f_{*} \in S_{F,x_{*}}\). Hence, \({\mathcal{A}}_{2}\) has a closed graph (and therefore has closed values). In consequence, the operator \({\mathcal{A}}_{2}\) is upper semicontinuous.
Thus, the operators \({\mathcal{A}}_{1}\) and \({\mathcal{A}}_{2}\) satisfy all the conditions of Theorem 4.1, and hence its conclusion implies that either condition (i) or condition (ii) holds. We show that conclusion (ii) is not possible. If \(x \in\lambda{\mathcal{A}}_{1} (x)+\lambda{\mathcal{A}}_{2}(x)\) for \(\lambda\in (0,1)\), then there exists \(f\in S_{F,x}\) such that
$$\begin{aligned} x(t) =& \lambda \int_{0}^{t} \frac{(t-s)^{\alpha -1}}{\Gamma(\alpha)}f(s)\,ds+\lambda \frac{2(1-t)}{\eta(2-\eta)}\xi +\lambda\frac{2t-\eta}{2-\eta}h(x) \\ &{}+\lambda \frac{2(t-1)}{\eta(2-\eta)} \int_{0}^{\eta}\frac{(\eta-s)^{\alpha }}{\Gamma(\alpha+1)}f(s)\,ds,\quad t \in[0,1]. \end{aligned}$$
Following the method of the proof of Claim I, we can obtain
$$\begin{aligned} \|x\| \le& \|p\| \bigl[\psi_{1}\bigl(\Vert x\Vert _{X} \bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X}\bigr)\bigr] \biggl[\frac {1}{\Gamma(\alpha+1)}+\frac{2}{\eta\Gamma(\alpha+2)} \biggr] \\ &{}+\frac{2}{\eta}|\xi|+(2+\eta)\ell\|x\|_{X} \end{aligned}$$
and
$$\begin{aligned} \bigl\Vert {}^{\mathrm{c}}D^{\beta}x\bigr\Vert \le& \bigl[ \psi_{1}\bigl(\Vert x\Vert _{X}\bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X}\bigr)\bigr] \biggl\{ \frac{1}{\Gamma(\alpha-\beta+1)} +\frac{2}{\eta\Gamma(2-\beta)}\frac{1}{\Gamma(\alpha+2)} \biggr\} \\ &{}+\frac{2}{\Gamma(2-\beta)} \biggl(\frac{|\xi|}{\eta}+ \ell\|x\| _{X} \biggr). \end{aligned}$$
Consequently,
$$\begin{aligned} \|x\|_{X} =&\| x\|+\bigl\Vert {}^{\mathrm{c}}D^{\beta} x \bigr\Vert \\ \le& \|p\| \bigl[\psi_{1}\bigl(\Vert x\Vert _{X} \bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X}\bigr)\bigr] \biggl\{ \frac {1}{\Gamma(\alpha+1)}+\frac{2}{\eta\Gamma(\alpha+2)}+\frac {1}{\Gamma(\alpha-\beta+1)} \\ &{}+\frac{2}{\eta\Gamma(2-\beta)}\frac{1}{\Gamma(\alpha+2)} \biggr\} + \frac{2}{\eta}|\xi| \biggl(1+\frac{1}{\Gamma(2-\beta)} \biggr) \\ &{}+ \biggl(2+\eta+\frac{2}{\Gamma(2-\beta)} \biggr)\ell\|x\|_{X}. \end{aligned}$$
Thus,
$$ \|x\|_{X}\le\Lambda_{1}\|p\| \bigl[ \psi_{1}\bigl(\Vert x\Vert _{X}\bigr)+ \psi_{2}\bigl(\Vert x\Vert _{X}\bigr)\bigr] + \Lambda_{2}+ \ell\Lambda_{3}\|x\|_{X}. $$
(4.5)
If condition (ii) of Theorem 4.1 holds, then there exist \(\lambda\in (0, 1)\) and \(x\in\partial B_{r}\) with \(x = \lambda {\mathcal{N}}(x)\). Then, x is a solution of (1.1) with \(\|x\|_{X} = M\). Now, by inequality (4.5) we get
$$\frac{(1-\ell\Lambda_{3})M}{ \Lambda_{1}\|p\| [\psi_{1}(M)+ \psi_{2}(M)]+\Lambda_{2}} \le1, $$
which contradicts (4.2). Hence, \({\mathcal{N}}\) has a fixed point in \([0,1]\) by Theorem 4.1, and consequently problem (1.1) has a solution. This completes the proof. □
Example 4.1
Consider the fractional differential inclusion
$$ {}^{\mathrm{c}}D^{5/2}x(t)\in F\bigl(t, x(t), {}^{\mathrm{c}}D^{3/4}x(t)\bigr) $$
(4.6)
supplemented with the boundary conditions (3.8), where
$$\begin{aligned} F\bigl(t, x(t), {}^{\mathrm{c}}D^{3/4}x(t)\bigr) =& \biggl[ \frac{1}{\sqrt{225+t}} \biggl(x(t) +\tan^{-1}\bigl({}^{\mathrm{c}}D^{3/4}x(t) \bigr)+\frac{\pi}{2} \biggr), \\ &\frac{1}{4} \sin x(t) +\frac{{}^{\mathrm{c}}D^{3/4}x(t)}{9(1+{}^{\mathrm{c}}D^{3/4}x(t))}+\frac{1}{10} \biggr]. \end{aligned}$$
Clearly,
$$\bigl\vert F\bigl(t,x, {}^{\mathrm{c}}D^{3/4}x(t)\bigr)\bigr\vert \le p(t) \bigl(\psi_{1}\bigl(\vert x\vert \bigr)+\psi_{2} \bigl(\bigl\vert {}^{\mathrm{c}}D^{3/4}x(t)\bigr\vert \bigr)\bigr), $$
where \(p(t)=1/\sqrt{225+t}\), \(\psi_{1}(|x|)=|x|\), \(\psi_{2}(|x|)=\pi\). Using the data of Example 3.1 and condition (H3), we find that \(M>M_{1}\simeq6.272351\). Hence, the hypothesis of Theorem 4.2 is satisfied, which implies that the fractional differential inclusion (4.6) together with (3.8) has a solution on \([0, 1]\).