Using the Hausdorff measure of noncompactness and the multivalued fixed point theorem, we shall give existence results for the nonlocal initial value problem (1.1)(1.2). First, we list the hypotheses of our first main result.
 (HA):

The \(C_{0}\) semigroup \(\{T(t)\}_{t\geq0}\) generated by the linear operator A is equicontinuous. We denote
$$M=\sup_{t\in{J}} \bigl\{ \bigl\Vert T(t)\bigr\Vert \bigr\} . $$
 (Hg):

The nonlocal term \(g: C(J, X)\rightarrow X\) is continuous and compact, and there exists a constant \(N>0\) such that \(\g(u)\\leq N\) for all \(u\in{C(J,X)}\).
 (HF):

The multivalued operator \(F: J\times X\rightarrow P_{cp,cv}(X)\) satisfies the following hypotheses:

(1)
F is measurable to t for every \(x\in X\) and u.s.c. to x for a.e. \(t\in J\). For every \(u \in C(J, X)\), the set
$$S_{F}(u)= \bigl\{ f\in L^{1}(J, X): f(t)\in F\bigl(t,u(t) \bigr), \mbox{ a.e. }t\in J \bigr\} $$
is nonempty.

(2)
There exists a function \(m\in L^{{1}/{q_{1}}}(J, \mathbb {R}^{+})\) with \(q_{1}\in(0, q)\) such that, for any \(x \in X\),
$$ \bigl\Vert F(t,x)\bigr\Vert =\sup \bigl\{ \Vert y\Vert : y \in F(t,x) \bigr\} \leq m(t) $$
(3.1)
for a.e. \(t\in J\).

(3)
There exists a constant \(L>0\) such that, for any bounded subset D of X, we have
$$ \beta\bigl(F(t, D)\bigr) \leq L\beta(D) $$
(3.2)
for a.e. \(t\in J\).
 (Hη):

There exists a function \(\eta\in C(J, \mathbb {R}^{+})\) such that, for each constant \(\lambda\in(1,0)\), we have
$$ \eta(t)\geq MN+\frac{qM}{\Gamma(1+q)}\frac{t^{(1+\lambda )(1q_{1})}}{(1+\lambda)^{(1q_{1})}}\m \_{L^{{1}/{q_{1}}}[0,t]},\quad t\in J, $$
(3.3)
where \(M, N, q_{1}\), and m are from (HA), (Hg), and (HF)(2).
In the proof of the existence results, we also need the following lemmas.
Lemma 8
[38]
Let
X
be a Banach space, and
F
be a multivalued mapping satisfying (HF)(1). Let
\(\mathcal{F}: L^{1}(J, X)\rightarrow C(J, X)\)
be a linear continuous mapping. Then the operator
$$\mathcal{F}\circ S_{F}: C(J, X)\rightarrow P_{cl,cv}\bigl(C(J, X)\bigr), $$
where
\((\mathcal{F}\circ S_{F})(u) =\mathcal{F}(S_{F}(u))\), is a closed graph operator in
\(C(J, X)\times C(J, X)\).
Lemma 9
[39] (Fixed point theorem)
If
W
is a bounded, closed, convex, and compact nonempty subset of
X
and the map
\(F: W\rightarrow2^{W}\)
is upper semicontinuous with
\(F(x)\)
being a closed and convex nonempty subset of
W
for each
\(x\in W\), then
F
has at least one fixed point in
W.
Lemma 10
[40]
Suppose that
\(b\geq0\), \(\sigma>0\), and
\(a(t)\)
is a nonnegative function locally integrable on
\(0\leq t< b\), and suppose that
\(c(t)\)
is nonnegative and locally integrable on
\(0\leq t< b\)
with
$$c(t)\leq a(t)+b \int_{0}^{t}(ts)^{\sigma1}c(s)\,\mathrm{d}s, \quad t\in[0,b). $$
Then
$$c(t)\leq a(t)+\mu \int_{0}^{t} E_{\sigma}^{\prime}\bigl( \mu(ts)\bigr)a(s) \,\mathrm{d}s,\quad t\in[0,b), $$
where
$$\mu=\bigl(b\Gamma(\sigma)\bigr)^{{1}/{\sigma}},\qquad E_{\sigma}(z)=\sum _{n=0}^{\infty} \frac{z^{n\sigma}}{\Gamma (n\sigma +1)},\qquad E_{\sigma}^{\prime}(z)=\frac{d}{dz}E_{\sigma}(z). $$
Now we are ready to prove the following existence result for the nonlocal fractional differential inclusion (1.1)(1.2).
Theorem 1
If hypotheses (HA), (Hg), (HF)(1)(2)(3), and (Hη) are satisfied, then the fractional differential inclusion (1.1)(1.2) has at least one mild solution on
J.
Proof
In view of (HF)(1), for each \(u \in C(J, X)\), the set \(S_{F}(u)\) is nonempty. So we can define the operator \(R: C(J, X)\rightarrow2^{C(J, X)}\) by
$$R(u)= \biggl\{ v\in C(J, X): v(t)=\mathcal{T}_{q}(t)g(u)+ \int _{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts) f(s)\,\mathrm{d}s, f\in S_{F}(u) \biggr\} . $$
Then, obviously, the fixed point of the operator R is a mild solution of (1.1)(1.2). So we shall show that R satisfies all the conditions of Lemma 9. For convenience, we subdivide the proof into four steps.
Step 1. We show that the values of R are convex and closed subsets in \(C(J, X)\).
We first prove that \(R(u)\) is convex for arbitrary \(u \in C(J, X)\). Indeed, if \(v_{1}\) and \(v_{2}\) belong to \(R(u)\), then there exist \(f_{1}\) and \(f_{2}\) belonging to \(S_{F}(u)\) such that, for every \(t\in J\) and \(i=1, 2\),
$$v_{i}(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f_{i}(s)\,\mathrm{d}s. $$
Let \(\lambda\in[0,1]\). Then, for every \(t\in J\) and \(i=1, 2\), we have
$$\lambda v_{1}(t)+(1\lambda)v_{2}(t)=\mathcal{T}_{q}(t)g(u)+ \int _{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts) \bigl(\lambda f_{1}(s)+(1 \lambda)f_{2}(s)\bigr)\,\mathrm{d}s. $$
It is easy to see that \(S_{F}(u)\) is convex since F has convex values. So \(\lambda f_{1}+(1\lambda)f_{2}\in S_{F}(u)\), and thus \(\lambda v_{1}+(1\lambda)v_{2}\in R(u)\). Therefore, \(R(u)\) is convex.
We now prove that \(R(u)\) is closed for every \(u\in C(J, X)\). To this end, suppose that \(\{v_{n}\}_{n=1}^{\infty}\) is a sequence in \(R(u)\) such that \(v_{n}\rightarrow v\) as \(n\rightarrow\infty\). Then we need to show that \(v \in R(u)\). According to the definition of R, there exists a sequence \(\{f_{n}\} _{n=1}^{\infty}\subset S_{F}(u)\) such that, for every \(t\in J\), we have
$$ v_{n}(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f_{n}(s)\,\mathrm{d}s. $$
(3.4)
For any \(n\geq1\) and almost all \(s\in(0,t]\), using assumption (HF)(2), we obtain
$$\bigl\ f_{n}(s)\bigr\ \leq m(s). $$
This shows that \(\{f_{n}:n\geq1\}\) is integrably bounded. Moreover, \(\{ f_{n}(t):n\geq1\} \subset F(t,u(t))\) implies that \(\{f_{n}(t):n\geq1 \}\) is relatively compact in X for a.e. \(t\in J\). Therefore, the set \(\{f_{n}:n\geq1\}\) is semicompact. By Lemma 6 it is weakly compact in \(L^{1}(J, X)\). We can assume that \(f_{n}\) converges weakly to a function \(f\in L^{1}(J, X)\). Then by Mazur’s lemma there is a sequence \(\{g_{n}\}_{n=1}^{\infty }\subseteq\overline{\operatorname{conv}}\{f_{n}:n\geq1\}\) such that \(g_{n}\) converges strongly to f. Since the values of F are convex and compact, we have that the set \(S_{F}(u)\) is convex and compact. So \(\{g_{n}\}_{n=1}^{\infty} \subseteq S_{F}(u)\), and \(f\in S_{F}(u)\).
Moreover, for all \(t \in J, s\in(0,t]\) and \(n\geq1\), we have
$$\bigl\Vert (ts)^{q1}\mathcal{S}_{q}(ts)f_{n}(s) \bigr\Vert \leq (ts)^{q1}\frac{Mq}{\Gamma(1+q)}m(s). $$
Let \(\lambda=({q1})/({1q_{1}})\in(1,0)\). Using Hölder’s inequality, we get
$$\begin{aligned} \int_{0}^{t}(ts)^{q1}m(s)\,\mathrm{d}s & \leq \biggl( \int_{0}^{t}(ts)^{\frac{q1}{1q_{1}}}\,\mathrm{d}s \biggr)^{1q_{1}} \biggl( \int_{0}^{t} \bigl(m(s) \bigr)^{\frac{1}{q_{1}}}\, \mathrm{d}s \biggr)^{q_{1}} \\ &\leq \frac{b^{(1+\lambda)(1q_{1})}}{(1+\lambda)^{(1q_{1})}}\m\ _{L^{{1}/{q_{1}}}[0,b]}< +\infty. \end{aligned}$$
Therefore, according to the Lebesgue dominated convergence theorem, taking \(n\rightarrow\infty\) on both sides of (3.4), we get
$$ v(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f(s)\,\mathrm{d}s, $$
(3.5)
which means that \(v\in R(u)\).
Step 2. We construct a bounded, convex, closed, and compact nonempty subset \(W\subseteq C(J, X)\) such that R maps W into itself.
Let \(W_{0}=\{u\in C(J, X): \u(t)\\leq\eta(t), t\in J\}\). It is obvious that \(W_{0}\subseteq C(J, X)\) is bounded, closed, and convex. We claim that \(R(W_{0})\subseteq W_{0}\). To prove this, let \(u\in W_{0}\) and \(v \in R(u)\). Then there exists \(f \in S_{F}(u)\) such that
$$v(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f(s)\,\mathrm{d}s. $$
According to (2.10) and assumptions (HF)(2) and (Hη), for every \(t\in J\), we have
$$\begin{aligned} \bigl\ v(t)\bigr\ &\leq\bigl\ \mathcal{T}_{q}(t)g(u)\bigr\ +\biggl\Vert \int _{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts)f(s)\,\mathrm{d}s\biggr\Vert \\ &\leq MN+\frac{qM}{\Gamma(1+q)} \int_{0}^{t}(ts)^{q1}m(s)\,\mathrm{d}s \\ &\leq MN+\frac{qM}{\Gamma(1+q)} \frac{t^{(1+\lambda)(1q_{1})}}{(1+\lambda)^{(1q_{1})}}\m\ _{L^{{1}/{q_{1}}}[0,t]} \\ &\leq \eta(t), \end{aligned}$$
which implies that \(R(W_{0})\subseteq W_{0}\).
Moreover, we claim that \(R(W_{0})\) is equicontinuous on J. For \(0\leq t < t+h \leq b\) and any \(v\in R(W_{0})\), there exist \(u \in W_{0}\) and \(f \in S_{F}(u)\) such that
$$\begin{aligned} \bigl\ v(t+h)v(t)\bigr\ \leq{}& \bigl\Vert \mathcal{T}_{q}(t+h)g(u) \mathcal{T}_{q}(t)g(u)\bigr\Vert +\biggl\Vert \int_{0}^{t+h}(t+hs)^{q1} \mathcal{S}_{q}(t+hs)f(s)\,\mathrm{d}s \\ &{} \int_{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts)f(s)\,\mathrm{d}s\biggr\Vert \\ \leq{}& G_{1}+G_{2}+G_{3}+G_{4}, \end{aligned}$$
where
$$\begin{aligned} &G_{1}= \bigl\ \mathcal{T}_{q}(t+h)g(u)\mathcal{T}_{q}(t)g(u) \bigr\ , \\ &G_{2}=\biggl\Vert \int_{0}^{t} \bigl[(t+hs)^{q1}(ts)^{q1} \bigr]\mathcal {S}_{q}(t+hs)f(s)\,\mathrm{d}s\biggr\Vert , \\ &G_{3}=\biggl\Vert \int_{0}^{t}(ts)^{q1} \bigl[\mathcal {S}_{q}(t+hs)\mathcal {S}_{q}(ts)\bigr]f(s)\,\mathrm{d}s \biggr\Vert , \\ &G_{4}=\biggl\Vert \int_{t}^{t+h}(t+hs)^{q1} \mathcal {S}_{q}(t+hs)f(s)\,\mathrm{d}s\biggr\Vert . \end{aligned}$$
We now show that \(G_{i}\rightarrow0\) as \(h \rightarrow0\) for \(i=1,2,3,4\). First, by assumption (HA) and Lemma 1(2) we know that \(\{\mathcal {T}_{q}(t): t\in J\}\) is equicontinuous. Combining this with the compactness of g, we have that \(G_{1} \rightarrow0\) as \(h\rightarrow0\), uniformly for all \(u\in W_{0}\).
Next, for \(G_{3}\), from the equicontinuity of \(\{\mathcal{S}_{q}(t): t\in J\}\), we can conclude that
$$\int_{0}^{t}\bigl\ (ts)^{q1} \bigl[\mathcal {S}_{q}(t+hs)\mathcal {S}_{q}(ts)\bigr]f(s)\bigl\ \,\mathrm{d}s \rightarrow0 $$
as \(h\rightarrow0\), uniformly for all \(f\in S_{F}(u)\) and \(u\in W_{0}\). That is, \(G_{3} \rightarrow0\) as \(h\rightarrow0\).
Then for \(G_{2}\), we note that \(\lambda={(q1)}/{(1q_{1})}\in(1,0)\). So for \(s< t\leq t+h\), we have \((ts)^{\lambda}\geq(t+hs)^{\lambda}\). Applying Lemma 7 and noting that \(1q_{1} \in(0,1)\), we get
$$\bigl\vert \bigl[(t+hs)^{\lambda} \bigr]^{1q_{1}} \bigl[(ts)^{\lambda } \bigr]^{1q_{1}}\bigr\vert \leq \bigl[(ts)^{\lambda}(t+hs)^{\lambda} \bigr]^{1q_{1}}. $$
Accordingly, we have
$$\begin{aligned} \bigl\vert (t+hs)^{q1}(ts)^{q1}\bigr\vert &=\bigl\vert \bigl[(t+hs)^{\lambda } \bigr]^{1q_{1}} \bigl[(ts)^{\lambda} \bigr]^{1q_{1}}\bigr\vert \\ &\leq \bigl[(ts)^{\lambda}(t+hs)^{\lambda} \bigr]^{1q_{1}}. \end{aligned}$$
Therefore, by (3.1), Lemma 1(1), and Lemma 7, we get
$$\begin{aligned} G_{2}&\leq \int_{0}^{t}\bigl\vert (t+hs)^{q1}(ts)^{q1} \bigr\vert \bigl\ \mathcal {S}_{q}(t+hs)f(s)\bigr\ \,\mathrm{d}s \\ &\leq \frac{Mq}{\Gamma(1+q)} \int_{0}^{t}\bigl\vert (t+hs)^{q1}(ts)^{q1} \bigr\vert m(s)\,\mathrm{d}s \\ &\leq \frac{Mq}{\Gamma(1+q)} \biggl( \int_{0}^{t}\bigl\vert (t+hs)^{q1}(ts)^{q1} \bigr\vert ^{\frac{1}{1q_{1}}}\,\mathrm{d}s \biggr)^{1q_{1}} \m \_{L^{{1}/{q_{1}}}[0,b]} \\ &\leq \frac{Mq}{\Gamma(1+q)} \biggl( \int_{0}^{t} \bigl[(ts)^{\lambda }(t+hs)^{\lambda} \bigr]\,\mathrm{d}s \biggr)^{1q_{1}} \m\_{L^{{1}/{q_{1}}}[0,b]} \\ &\leq \frac{Mq}{(1+\lambda)^{(1q_{1})}\Gamma(1+q)} \bigl[h^{1+\lambda }+t^{1+\lambda}(t+h)^{1+\lambda} \bigr]^{1q_{1}} \m\_{L^{{1}/{q_{1}}}[0,b]} \\ &\leq \frac{Mq}{(1+\lambda)^{(1q_{1})}\Gamma(1+q)}h^{(1+\lambda)(1q_{1})} \m\_{L^{{1}/{q_{1}}}[0,b]}. \end{aligned}$$
Therefore, \(G_{2} \rightarrow0\) as \(h\rightarrow0\).
Finally, for \(G_{4}\), we have
$$\begin{aligned} G_{4}&\leq \int_{t}^{t+h}\bigl\vert (t+hs)^{q1}\bigr\vert \bigl\ \mathcal {S}_{q}(t+hs)f(s)\bigr\ \,\mathrm{d}s \\ &\leq \frac{Mq}{\Gamma(1+q)} \int_{t}^{t+h}(t+hs)^{q1}m(s) \,\mathrm{d}s \\ &\leq \frac{Mq}{\Gamma(1+q)} \biggl( \int_{t}^{t+h} \bigl[(t+hs)^{q1} \bigr]^{\frac{1}{1q_{1}}}\,\mathrm{d}s \biggr)^{1q_{1}} \m\_{L^{{1}/{q_{1}}}[t,t+h]} \\ &\leq \frac{Mq}{\Gamma(1+q)(1+\lambda)^{(1q_{1})}}h^{(1+\lambda)(1q_{1})} \m\_{L^{{1}/{q_{1}}}[t,t+h]}. \end{aligned}$$
So we also have that \(G_{4} \rightarrow0\) as \(h\rightarrow0\). Therefore, we get \(\v(t+h)v(t)\\rightarrow0\) as \(h\rightarrow0\), implying that \(R(W_{0})\) is equicontinuous on J.
Now, let us define a sequence \(\{W_{n}\}_{n=1}^{\infty}\) recursively by \(W_{n}=\overline{\operatorname{conv}} R(W_{n1})\) for all \(n\geq1\). From the above discussions we know that \(W_{n}\subseteq W_{n1}\) for all \(n\geq1\). Thus, \(\{W_{n}\}_{n=1}^{\infty}\) is a decreasing sequence of closed, bounded, convex, and equicontinuous subsets of \(C(J, X)\). Set \(W=\bigcap_{n=1}^{\infty}W_{n}\). Then W is a closed, bounded, convex, and equicontinuous subset of \(C(J, X)\) and \(R(W)\subseteq W\). We claim that W is nonempty and compact in \(C(J,X)\). Indeed, by Lemma 2 and (2.15), for arbitrary \(\varepsilon>0\), there exist sequences \(\{u_{k}\}_{k=1}^{\infty}\subset W_{n}\) and \(\{f_{k}\}_{k=1}^{\infty }\subset S_{F}(u_{k})\) such that
$$\begin{aligned} \beta\bigl(W_{n+1}(t)\bigr)&=\beta\bigl(\overline{\operatorname{conv}}(R W_{n}) (t)\bigr) = \beta\bigl((R W_{n}) (t)\bigr) \\ &\leq 2\beta\bigl(\bigl\{ R u_{k}(t)\bigr\} _{k=1}^{\infty} \bigr)+\varepsilon \\ &\leq 2\beta \biggl(\mathcal{T}_{q}(t)g\bigl(\{u_{k} \}_{k=1}^{\infty}\bigr)+ \int _{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts)\bigl\{ f_{k}(s)\bigr\} _{k=1}^{\infty }\,\mathrm{d}s \biggr)+\varepsilon \\ &\leq 2\beta \bigl(\mathcal{T}_{q}(t)g\bigl(\{u_{k} \}_{k=1}^{\infty}\bigr) \bigr)+2\beta \biggl( \int_{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts)\bigl\{ f_{k}(s)\bigr\} _{k=1}^{\infty}\,\mathrm{d}s \biggr)+\varepsilon. \end{aligned}$$
Since g is compact, by Lemma 2, (2.14), and (3.2) we have
$$\begin{aligned} \beta\bigl(W_{n+1}(t)\bigr)&\leq 4 \int_{0}^{t}(ts)^{q1}\beta \bigl( \mathcal{S}_{q}(ts)\bigl\{ f_{k}(s)\bigr\} _{k=1}^{\infty} \bigr)\,\mathrm{d}s+\varepsilon \\ &\leq \frac{4qM}{\Gamma(1+q)} \int_{0}^{t}(ts)^{q1}\beta \bigl(F \bigl(s,W_{n}(s) \bigr) \bigr)\,\mathrm{d}s+\varepsilon \\ &\leq \frac{4qML}{\Gamma(1+q)} \int_{0}^{t}(ts)^{q1} \beta \bigl(W_{n}(s)\bigr)\,\mathrm{d}s+\varepsilon. \end{aligned}$$
(3.6)
Since (3.6) is true for arbitrary \(\varepsilon>0\), we must have
$$ \beta\bigl(W_{n+1}(t)\bigr)\leq \frac{4qML}{\Gamma(1+q)} \int_{0}^{t}(ts)^{q1} \beta \bigl(W_{n}(s)\bigr)\,\mathrm{d}s. $$
(3.7)
Because \(\{W_{n}\}_{n=1}^{\infty}\) is decreasing with respect to n, we define
$$\mu(t) =\lim_{n\rightarrow\infty} \beta\bigl(W_{n}(t)\bigr),\quad t\in J. $$
Taking \(n\rightarrow\infty\) on both sides of (3.7), we have
$$\mu(t) \leq\frac{4qML}{\Gamma(1+q)} \int_{0}^{t}(ts)^{q1} \mu (s)\, \mathrm{d}s. $$
By Lemma 10 we have that \(\mu(t) =0, t\in J\). On the other hand, \(\{W_{n}\}_{n=1}^{\infty}\) is bounded and equicontinuous. Hence, by Lemma 3 we get \(\beta(W_{n}) =\sup_{t\in J}\beta(W_{n}(t))\), which implies that \(\lim_{n\rightarrow\infty}\beta(W_{n})=0\). Then it follows from Lemma 2(8) that \(W=\bigcap_{n=1}^{\infty}W_{n}\) is nonempty and compact in \(C(J, X)\).
Step 3. We show that the graph of R is closed.
Let \(\{u_{n}\}_{n=1}^{\infty}\subset W\) with \(u_{n}\rightarrow u\) as \(n\rightarrow\infty\), \(v_{n}\in R (u_{n})\), and \(v_{n}\rightarrow v\) as \(n\rightarrow\infty\). We shall show that \(v \in R(u)\). Since \(v_{n}\in R (u_{n})\), there exists \(f_{n}\in S_{F}(u_{n})\) such that
$$v_{n}(t)=\mathcal{T}_{q}(t)g(u_{n})+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f_{n}(s)\,\mathrm{d}s. $$
In the sequel, we must show that there exists \(f\in S_{F}(u)\) such that
$$v(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f(s)\,\mathrm{d}s. $$
Consider the linear operator \(\mathcal{F}: L^{1}(J, X)\rightarrow C(J, X)\) defined as
$$\mathcal{F}(f) (t)= \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f(s)\,\mathrm{d}s. $$
Clearly, \(\mathcal{F}\) is linear and continuous. Hence, it follows from Lemma 8 that \(\mathcal{F}\circ S_{F}\) is a closed graph operator. Moreover, we have
$$v_{n}(\cdot)\mathcal{T}_{q}(\cdot)g(u_{n}) \in \mathcal{F}\bigl(S_{F}(u_{n})\bigr). $$
Since \(v_{n}\rightarrow v\), \(u_{n}\rightarrow u\), and g is continuous, we obtain
$$v(\cdot)\mathcal{T}_{q}(\cdot)g(u)\in\mathcal{F} \bigl(S_{F}(u)\bigr), $$
that is,
$$v(t)\mathcal{T}_{q}(t)g(u)= \int_{0}^{t}(ts)^{q1} \mathcal {S}(ts)f(s)\,\mathrm{d}s $$
for some \(f\in S_{F}(u)\), and thus the graph of R is closed.
Step 4. We show that R is u.s.c. on W.
As a consequence of the previous proof, we have that W is closed and \(R(u)\) is closed for all \(u\in W\). The set \(\overline{R(W)}\subseteq W\) is compact. Moreover, R is closed. According to Remark 1, we can come to the conclusion that R is u.s.c.
Finally, due to the fixed point of Lemma 9, R has at least one point \(u \in R (u)\), and u is a mild solution to the fractional semilinear differential inclusion (1.1) with the nonlocal condition (1.2). □
Remark 2
If A generates an analytic semigroup or a differential semigroup \(\{ T(t)\}_{t\geq0}\), then \(\{T(t)\}_{t\geq0}\) is equicontinuous [41]. In applications of partial differential equations, such as parabolic equations and strongly damped wave equations, the corresponding solution semigroups are analytic. Therefore, the results in this paper have wide applicability.
In Theorem 1, hypothesis (HF)(2) may be sometimes difficult to be verified, and also the mapping g needs to be uniformly bounded. Indeed, if g is compact, then g is bounded on a bounded subset. Next, we further give an existence result for the fractional differential inclusion (1.1)(1.2) under a relatively weak condition of F when g is not uniformly bounded. For this purpose, we introduce the following two hypotheses:
 (Hg′):

The nonlocal term \(g: C(J, X)\rightarrow X\) is continuous and compact.
 (HF)(2′):

There exist a function \(\alpha\in L^{{1}/{q_{1}}}(J, \mathbb{R}^{+})\) for some given \(q_{1}\in(0,q)\) and an increasing function \(\Omega :\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) such that
$$\bigl\Vert F(t,x)\bigr\Vert \leq\alpha(t)\Omega\bigl(\Vert x\Vert \bigr) $$
for a.e. \(t\in J\) and all \(x\in X\).
Theorem 2
Assume that hypotheses (HA), (Hg′), and (HF)(1)(2′)(3) are satisfied. Suppose that
$$ \lim_{k\rightarrow\infty} \sup \biggl\{ \frac{M}{k} \biggl(\gamma (k)+\frac{M_{1}q}{(1+\lambda)^{(1q_{1})} \Gamma(1+q)}\Omega (k)b^{(1+\lambda)(1q_{1})} \biggr) \biggr\} < 1, $$
(3.8)
where
\(\gamma(k)=\sup \{\g(u)\: \u\ \leq k \}\), \(M_{1}=\ \alpha\_{L^{{1}/{q_{1}}}[0,b]}\), and
\(\lambda=({q1})/({1q_{1}})\in(1,0)\). Then the fractional differential inclusion (1.1)(1.2) has at least one mild solution on
J.
Proof
By (3.8) there exists a constant \(k>0\) such that
$$M \biggl(\gamma(k)+\frac{M_{1}q}{(1+\lambda)^{(1q_{1})} \Gamma (1+q)}\Omega(k)b^{(1+\lambda)(1q_{1})} \biggr)< k. $$
Let \(W_{0}=\{u \in C(J, X): \u\\leq k\}\) and \(W_{1}=\overline{\operatorname{conv}} R(W_{0})\). Then for any \(v \in W_{1}\), there exist \(u\in W_{0}\) and \(f \in S_{F}(u)\) such that
$$v(t)=\mathcal{T}_{q}(t)g(u)+ \int_{0}^{t}(ts)^{q1}\mathcal {S}_{q}(ts)f(s)\,\mathrm{d}s. $$
From (2.10) and assumption (Hg′) we conclude that, for every \(t\in J\), we have
$$\begin{aligned} \bigl\ v(t)\bigr\ &\leq\bigl\Vert \mathcal{T}_{q}(t)g(u)\bigr\Vert +\biggl\Vert \int _{0}^{t}(ts)^{q1} \mathcal{S}_{q}(ts)f(s)\,\mathrm{d}s\biggr\Vert \\ &\leq M \biggl(\gamma(k)+\frac{q}{\Gamma(1+q)}\Omega(k) \int _{0}^{t}(ts)^{q1}\alpha(s)\, \mathrm{d}s \biggr) \\ &\leq M \biggl(\gamma(k)+\frac{q}{\Gamma(1+q)}\Omega(k) \biggl( \int _{0}^{t}(ts)^{\frac{q1}{1q_{1}}}\,\mathrm{d}s \biggr)^{(1q_{1})} \\alpha \_{L^{{1}/{q_{1}}}[0,b]} \biggr) \\ &\leq M \biggl(\gamma(k)+\frac{M_{1}q}{(1+\lambda)^{(1q_{1})} \Gamma (1+q)}\Omega(k)b^{(1+\lambda)(1q_{1})} \biggr) \\ &< k. \end{aligned}$$
It follows that \(\v\=\sup_{t\in J} \{\v(t)\ \}< k\). Then \(W_{1}\subseteq W_{0}\). Set \(W_{n}=\overline{\operatorname{conv}}R(W_{n1})\) for all \(n\geq1\). Then we can complete the proof similarly to that of Theorem 1. □