## 1 Introduction

A lot of papers on fractional differential equations (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18] and the references therein) have been published. As you know, most famous fractional derivations are the Caputo and Riemann–Liouville derivations. In 2015, Caputo and Fabrizio introduced a new fractional derivation without singular kernel [19]. Some researchers published some works about solving different equations including the new derivation (see, for example, [2, 3, 10, 20,21,22,23,24,25]). Some researchers investigated some results on dimension of the set of solutions for some fractional differential inclusions (see, for example, [26]).

Let $$b>0$$, $$u\in H^{1}(0,b)$$, and $$\zeta \in (0,1)$$. As you know, the Caputo–Fabrizio fractional derivative of order ζ is defined by

$${}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)=\frac{(2-\zeta )M(\zeta )}{2(1-\zeta )} \int _{0}^{t}\exp \biggl(\frac{-\zeta }{1-\zeta }(t-s) \biggr)u^{\prime }(s)\,ds,$$

where $$t\geq 0$$ and $$M(\zeta )$$ is a normalization constant depending on ζ such that $$M(0)= M(1)=1$$ [19]. Losada and Nieto showed that $${}^{\mathrm{CF}}\mathcal{I}^{\zeta } u(t)=\frac{2(1-\zeta )}{(2- \zeta )M(\zeta )}u(t) +\frac{\zeta }{(2-\zeta )M(\zeta )}\int _{0}^{t} u(s) \,ds$$ [27]. Also, they showed that $$M(\zeta )=\frac{2}{2- \zeta }$$ [27]. Hence, the fractional Caputo–Fabrizio derivative of order ζ is given by $${}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)=\frac{1}{1- \zeta }\int _{0}^{t}\exp (-\frac{\zeta }{1-\zeta }(t-s))u^{\prime }(s)\,ds$$, when $$t\geq 0$$ and $$0<\zeta <1$$ [27]. If $$n\geq 1$$ and $$\zeta \in (0,1)$$, then the fractional derivative $${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta +n}$$ of order $$n+\zeta$$ is defined by $${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta +n}u:={}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }({\mathcal{D}}^{n}u(t))$$ [27]. Let $$u,v\in H^{1}(0,1)$$ and $$\zeta \in (0,1)$$. If $$u^{(s)}(0)=0$$ for all $$s=1,2,\ldots ,n$$, then $${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }({}^{\mathrm{CF}}{\mathcal{D}}^{n}(u(t))={}^{\mathrm{CF}}{\mathcal{D}}^{n}({}^{\mathrm{CF}} {\mathcal{D}}^{\zeta }(u(t))$$. Also, we have $$\lim_{\zeta \to 0}{}^{\mathrm{CF}} {\mathcal{D}}^{\zeta }u(t)=u(t)-u(0)$$, $$\lim_{\zeta \to 1}{}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }u(t)=u(t)^{\prime }$$, and $${}^{\mathrm{CF}}{\mathcal{D}}^{\zeta } ( \lambda u(t)+\gamma v(t) )=\lambda {}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)+ \gamma ^{\mathrm{CF}}{\mathcal{D}}^{\zeta }v(t)$$ [27]. It has been proved that the unique solution for the problem $${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }u(t)=v(t)$$ with boundary condition $$u(0)=c$$ is given by $$u(t)=c+a_{\zeta }(v(t)-v(0))+b_{\zeta }\int _{0}^{t} v(s)\,ds$$, where $$a_{\zeta } = \frac{2(1-\zeta )}{(2-\zeta )M(\zeta )}=1-\zeta$$ and $$b_{\zeta } = \frac{2\zeta }{ (2-\zeta )M(\zeta )}=\zeta$$ ([19] and [27]). Note that $$v(0)=0$$. Suppose that $$u,v\in C_{\mathbb{R}}[0,1]$$, $$u(0)=0$$, and there is a real constant L such that $$|u(t)-v(t)|\leq L$$ for all $$t\in [0,1]$$. Recently, Baleanu, Mousalou, and Rezapour proved that $$|^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)-{^{\mathrm{CF}}}{\mathcal{D}}^{\zeta }v(t)| \leq \frac{1}{(1-\zeta )^{2}}L$$ for all $$t\in [0,1]$$ [10]. This leads to $$|^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)| \leq (\frac{1}{(1-\zeta )^{2}})L$$ for all $$t\in [0,1]$$ whenever $$u\in C_{\mathbb{R}}[0,1]$$ and $$|u(t)|\leq L$$ for some $$L\geq 0$$ and all $$t\in [0,1]$$ with $$u(0)=0$$ [10]. Also, they showed that $$|{}^{\mathrm{CF}}\mathcal{I}^{\zeta }u(t)-{}^{\mathrm{CF}}\mathcal{I}^{\zeta }v(t)| \leq L$$ for all $$t\in [0,1]$$ [10] and so $$|^{\mathrm{CF}}\mathcal{I}^{\zeta }u(t)| \leq L$$ for all $$t\in [0,1]$$ whenever $$u\in C_{\mathbb{R}}[0,1]$$ with $$|u(t)|\leq L$$ for some $$L\geq 0$$ and all $$t\in [0,1]$$. For some more necessary definitions, see [1].

Let $$u\in C_{\mathbb{R}}[0,d]$$, $$d>0$$ and $$\zeta \in (0,1)$$. The extended fractional Caputo–Fabrizio derivation of order ζ is defined by [11]

\begin{aligned} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t) ={}& \frac{B(\zeta )}{1- \zeta } \bigl(u(t)-u (0)\bigr)\exp \biggl(\frac{-\zeta }{1-\zeta }t\biggr) \\ & {} +\frac{\zeta B(\zeta )}{(1-\zeta )^{2}} \int _{0}^{t} \bigl(u(t)-u (s)\bigr) \exp \biggl( \frac{-\zeta }{1-\zeta }(t-s)\biggr) \,ds. \end{aligned}

If $$u(0)=0$$, then we have $${}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)=\frac{B( \zeta )}{1-\zeta }u(t) - \frac{\zeta B(\zeta )}{(1-\zeta )^{2}}\int _{0}^{t} \exp (-\frac{\zeta }{1-\zeta }(t-s))u(s)\,ds$$ [11].

### Lemma 1

([11])

Let $$u\in H^{1}(0,b)$$, $$b>0$$, and $$\zeta \in (0,1)$$. Then $${}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)={}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)$$. If $$u\in C_{\mathbb{R}}[0,b]$$, then $$\lim_{\zeta \to 0}~{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)=u(t)-u(0)$$.

### Lemma 2

([11])

Let $$0<\zeta <1$$. Then a solution for the problem $${}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta }u(t)=v(t)$$ with boundary condition $$u(0)=0$$ is given by $$u(t)= a_{\zeta } v(t) + b_{\zeta }\int _{0}^{t} v(s)\,ds$$.

### Lemma 3

([11])

Let $$u,v\in C_{\mathbb{R}}[0,1]$$. If there is a real constant L such that $$|u(t)-v(t)|\leq L$$ for all $$t\in [0,1]$$, then $$|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta }v(t)| \leq \frac{(2-\zeta )B(\zeta ) }{(1-\zeta )^{2}}L$$ for all $$t\in [0,1]$$. If $$u(0)=v(0)$$, then $$|{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta }u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }v(t)| \leq \frac{B( \zeta ) }{(1-\zeta )^{2}}L$$.

This result implies that $$|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)| \leq \frac{(2-\zeta )B(\zeta ) }{(1-\zeta )^{2}}L$$ for all $$t\in [0,1]$$ whenever $$u\in C_{\mathbb{R}}[0,1]$$ with $$|u(t)|\leq L$$ for some $$L\geq 0$$ and all $$t\in [0,1]$$.

We need the following results.

### Lemma 4

([28])

Suppose that $$\mathcal{Y}$$ is a Banach space, $${\mathcal{F}}: I\times \mathcal{Y}\to \mathcal{P}_{cp,cv}( \mathcal{Y})$$ is an $$L^{1}$$-Caratheodory multivalued and ϵ is a linear continuous mapping from $$L^{1}(I,\mathcal{Y})$$ to $$C(I, \mathcal{Y})$$. Then the mapping $$\epsilon \circ S_{{\mathcal{F}}} : C(I, \mathcal{Y})\to \mathcal{P}_{cp,cv}C( I,\mathcal{Y})$$ defined by $$(\epsilon \circ S_{{\mathcal{F}}}) (y) = \epsilon (S_{{ \mathcal{F}},y})$$ is a closed graph mapping in $$C(I,\mathcal{Y})\times C(I,\mathcal{Y})$$.

### Theorem 5

([29])

Assume that Y is a Banach space, D is a closed and convex subset of Y, and W is an open subset of D with $$0\in W$$. If $${\mathcal{F}}: \bar{W}\to P_{cp,c}(D)$$ is an upper semi-continuous compact map, then either $${\mathcal{F}}$$ has a fixed point in or there is $$x\in \partial W$$ and $$\delta \in (0,1)$$ such that $$x \in \delta {\mathcal{F}}(x)$$.

### Theorem 6

([30])

Suppose that $$(\mathcal{Y},d)$$ is a complete metric space. If $${\mathcal{G}}: \mathcal{Y} \to P_{cl}(\mathcal{Y})$$ is a contraction, then $${\mathcal{G}}$$ has a fixed point.

### Theorem 7

([31])

Assume that $$\mathcal{Y}$$ is a Banach space, $$\mathcal{E}\in P_{bd,cl,cv}(\mathcal{Y})$$ and $$\mathcal{F}, \mathcal{G}: \mathcal{E}\to P_{cp,cv}(\mathcal{Y})$$ are two multivalued operators. If $$\mathcal{F}y+\mathcal{G}y\subset \mathcal{E}$$ for all $$y\in \mathcal{E}$$, $$\mathcal{F}$$ is a contraction and $$\mathcal{G}$$ is an upper semi-continuous compact map, then there is $$y\in \mathcal{E}$$ such that $$y\in \mathcal{F}y+\mathcal{G}y$$.

### Theorem 8

([32])

Assume that $$\mathcal{Y}$$ is a Banach algebra, $$D\in \mathcal{P}_{bd,cl,cv}(\mathcal{Y})$$ and $${\mathcal{F}_{1}}: D \to \mathcal{P}_{cl,cv,bd} (\mathcal{Y})$$ and $${\mathcal{F}_{2}}: D \to \mathcal{P}_{cp,cv} (\mathcal{Y})$$ are two set-valued maps such that $${\mathcal{F}_{1}}$$ is Lipschitz with a Lipschitz constant δ, $${\mathcal{F}}_{2}$$ is upper semi-continuous and compact, $${\mathcal{F} _{1}} x{\mathcal{F}_{2}}x$$ is a convex subset D for all $$x\in D$$ and $$\mathcal{N}\delta <1$$, where $$\mathcal{N} =\|{\mathcal{F}_{2}}(D)\|= \sup \{\|{\mathcal{F}_{2}}x\|: x\in D\}$$. Then there is $$y\in D$$ such that $$y\in {\mathcal{F}_{1}}y{\mathcal{F}_{2}}y$$.

### Lemma 9

([26])

Let $$\mathcal{A}$$ mapping $$[0,1]$$ into $$\mathcal{P}_{cp,cv} (\mathbb{R})$$ be measurable such that the Lebesgue measure of the set $$\{t: \dim \mathcal{A}(t)<1\}$$ is zero. Then there are arbitrarily many linearly independent measurable selections $$y_{1}(\cdot),\ldots , y_{m}(\cdot)$$ of $$\mathcal{A}$$.

### Theorem 10

([26])

Let $$\mathcal{H}$$ be a nonempty closed convex subset of a Banach space $$\mathcal{Y}$$ and $$\mathcal{F}: \mathcal{H} \to \mathcal{P}_{cp,cv}(\mathcal{H})$$ be a δ-contraction. If $$\dim \mathcal{F}(x)\geq m$$ for all $$x\in \mathcal{H}$$, then $$\dim Fix (\mathcal{F})\geq m$$.

## 2 Main results

Consider the Banach space $$\mathcal{X}=C(I)$$ of real-valued continuous functions on $$I=[0,1]$$ via the norm $$\|x\|=\sup_{t\in I}|x(t)|$$. Assume that $$\zeta ,\iota :[0,1] \times [0,1]\to [0,\infty )$$ are two continuous maps such that $$\sup |\int _{0}^{t} \iota (t,s) \,ds|<\infty$$ and $$\sup |\int _{0}^{t} \zeta (t,s) \,ds|<\infty$$. Consider the maps ϕ and φ defined by $$(\phi w)(t)= \int _{0}^{t} \zeta (t,s)w(s)\,ds$$ and $$(\varphi w)(t)= \int _{0}^{t} \iota (t,s)w(s)\,ds$$. Suppose that $$\eta (t)\in L^{\infty }(I)$$ with $$\eta ^{\ast }= \sup_{t\in I} |\eta (t)|$$. Put $$\zeta _{0}=\sup |\int _{0}^{t} \zeta (t,s) \,ds|$$ and $$\iota _{0}=\sup |\int _{0}^{t} \iota (t,s) \,ds|$$. First, we are going to investigate the fractional integro-differential inclusion

$${}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }x(t) \in \mathcal{F}\bigl(t, x(t),( \phi x) (t), (\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{m}}x(t)\bigr),$$
(1)

with boundary condition $$x(0)=0$$, where $$\zeta ,\beta _{1},\dots , \beta _{m}\in (0,1)$$.

We say that a function $$x\in \mathcal{X}$$ is a solution for problem (1) whenever there exists a function $$f\in C(I)$$ such that

$$f(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) ,\ldots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{m}}x(t) \bigr)$$

for almost all $$t\in I$$ and $$x(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0} ^{t} f(s)\,ds$$.

### Theorem 11

Let $$\mathcal{F}: I\times \mathbb{R}^{m+3}\to P_{cp,cv}(\mathbb{R})$$ be a Caratheodory multivalued map such that

\begin{aligned} \bigl\Vert \mathcal{F}(t,x_{1},x_{2},x_{3},y_{1}, \ldots , y_{m}) \bigr\Vert _{p} & = \sup \bigl\{ \vert y \vert : y \in \mathcal{F} (t,x_{1}, x_{2}, x_{3}, y_{1}, \ldots , y_{m})\bigr\} \\ & \leq \eta (t) \Biggl( \vert x_{1} \vert + \vert x_{2} \vert + \vert x_{3} \vert + \sum _{i=1}^{m} \vert y_{i} \vert \Biggr) \end{aligned}

for all $$t\in I$$, $$x_{i}, y_{j} \in \mathbb{R}$$, $$1\leq i\leq 3$$ and $$1\leq j \leq m$$. If $$\eta ^{*}(1+\zeta _{0} + \iota _{0} +\sum_{i=1} ^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )\leq 1$$, then inclusion (1) has one solution.

### Proof

For $$x\in \mathcal{X}$$, define a selection set of $$\mathcal{F}$$ at $$x\in \mathcal{X}$$ by

\begin{aligned} \begin{aligned} S_{\mathcal{F},x} :={} & \bigl\{ f\in L^{1} (I,R): f(t) \in \mathcal{F} \bigl(t,x(t),( \phi x) (t),(\varphi x) (t), \\ & {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}} _{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \beta _{m}}x(t) \bigr) \text{ for all } t\in I\bigr\} . \end{aligned} \end{aligned}

Since $$\mathcal{F}$$ is a Caratheodory multifunction, by using Theorem 1.3.5 in [33], we get $$S_{\mathcal{F},x}$$ is nonempty. Define an operator $$\varOmega \colon \mathcal{X} \to P( \mathcal{X})$$ by $$\varOmega (x)=\{g\in \mathcal{X}:\mbox{ there exists } f \in S_{\mathcal{F},x} \mbox{ such that } g(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0}^{t} f(s)\,ds\mbox{ for all }t\in I\}$$. We show that the operator Ω satisfies the hypothesis of Theorem 5. First, we show that $$\varOmega (x)$$ is convex for all $$\in \mathcal{X}$$.

Let $$g_{1},g_{2}\in \varOmega (x)$$ and $$w\in [0,1]$$. Choose $$f_{1},f_{2} \in S_{\mathcal{F},x}$$ such that $$g_{i}(t)=a_{\zeta }f_{i}(t)+b_{ \zeta }\int _{0}^{t} f_{i}(s)\,ds$$ for all $$t\in I$$. Then we have

$$\bigl[wg_{1}+(1-w)g_{2}\bigr](t)=a_{\zeta } \bigl(wf_{1}+(1-w)f_{2}\bigr) (t)+b_{\zeta } \int _{0}^{t}\bigl(wf_{1}+(1-w)f_{2} \bigr) (s)\,ds$$

for all $$t\in I$$. Since $$\mathcal{F}$$ has convex values, it is easy to check that $$S_{\mathcal{F},x}$$ is convex, and so $$wg_{1}+(1-w)g_{2} \in \varOmega (x)$$. Now, we show that Ω maps bounded sets into bounded subsets. Let $$\mathcal{B}_{r}=\{x\in \mathcal{X}:\|x\|\leq r \}$$, $$x\in \mathcal{B}_{r}$$, and $$g\in \varOmega (x)$$. Choose $$f\in S _{\mathcal{F},x}$$ such that

\begin{aligned} \bigl\vert g(t) \bigr\vert \leq{} &a_{\zeta } \bigl\vert f(t) \bigr\vert + b_{\zeta } \int _{0}^{t} \bigl\vert f(s) \bigr\vert \,ds \leq a_{\zeta }\eta (t) \bigl( \vert x \vert + \bigl\vert \varphi (x) \bigr\vert + \bigl\vert \phi (x) \bigr\vert \\ & {} + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}}x(t) \bigr\vert + \bigl\vert {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{2}}x(t) \bigr\vert + \cdots + \bigl\vert {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{m}}x(t) \bigr\vert \bigr) \\ & {} +b_{\zeta } \int _{0}^{t} \bigl( \vert x \vert + \bigl\vert \varphi (x) \bigr\vert + \bigl\vert \phi (x) \bigr\vert \\ & {} + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}}x(s) \bigr\vert + \bigl\vert {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{2}}x(s) \bigr\vert + \cdots + \bigl\vert {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{m}}x(s) \bigr\vert \bigr)\eta (s)\,ds \\ \leq{}& a_{\zeta } \eta ^{\ast } \Biggl(r+\zeta _{0} r+ \iota _{0} r+ \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}r \Biggr) \\ & {} + b_{\zeta } \eta ^{\ast } \Biggl( r+\zeta _{0} r+\iota _{0} r+ \sum_{i=1}^{m} \frac{B( \beta _{i})}{ (1-\beta _{i})^{2}}r \Biggr) \\ ={}& \eta ^{\ast }\cdot r\cdot \Biggl(1+\zeta _{0}+\iota _{0}+ \sum_{i=1}^{m} \frac{B( \beta _{i})}{( 1-\beta _{i})^{2}} \Biggr) (a_{\zeta }+b_{\zeta })\leq r. \end{aligned}

Thus, $$\|g\|=\max_{t\in I}| g(t) |\leq r$$. This implies that Ω maps bounded sets into bounded sets in $$\mathcal{X}$$. Now, we show that Ω maps bounded sets of $$\mathcal{X}$$ into equi-continuous sets. Let $$t_{1},t_{2}\in I$$ with $$t_{1}< t_{2}$$, $$x \in \mathcal{B}_{r}$$ and $$g\in \varOmega (x)$$. Then we have

\begin{aligned} \bigl\vert g(t_{2})-g(t_{1}) \bigr\vert & = \biggl\vert a_{\zeta }f(t_{2}) + b_{\zeta } \int _{0}^{t _{2}}f(s) \,ds - a_{\zeta }f(t_{1}) -b_{\zeta } \int _{0}^{t_{1}}f(s) \,ds \biggr\vert \\ & \leq a_{\zeta } \bigl\vert f(t_{2}) - f(t_{1}) \bigr\vert +b_{\zeta } \int _{t_{1}}^{t _{2}} \bigl\vert f(s) \bigr\vert \,ds \\ & \leq r \Biggl( 1+\zeta _{0} + \iota _{0}+\sum _{i=1}^{m}\frac{B(\beta _{i})}{(1 -\beta _{i})^{2}} \Biggr) \bigl(\eta (t_{2}) - \eta (t_{1}) \bigr) (a_{\zeta }+b_{\zeta }). \end{aligned}

Hence, the right-hand side of the inequality tends to zero (independent on $$x\in \mathcal{B}_{r}$$) as $$t_{2}\to t_{1}$$. This implies that $$\varOmega \colon \mathcal{X} \to P(\mathcal{X})$$ is a compact multivalued map by using the Arzela–Ascoli theorem. We show that Ω has a closed graph. Let $$x_{n} \to x_{\ast }$$, $$g_{n}\in \varOmega (x_{n})$$ for all n and $$g_{n} \to g_{\ast }$$. It is sufficient to prove that $$g_{\ast }\in \varOmega (x_{\ast })$$. Since $$g_{n}\in \varOmega (x_{n})$$ for all n, there exist $$f_{n}\in S_{\mathcal{F},x_{n}}$$ such that $$g_{n}(t)= a_{\zeta }f_{n}(t)+b_{\zeta } \int _{0}^{t} f_{n}(s)\,ds$$ for all $$t\in I$$. Thus, we have to show that there exist $$f_{\ast }\in S _{\mathcal{F},x_{\ast }}$$ such that $$g_{\ast }(t)=a_{\zeta }f_{\ast }(t)+b _{\zeta } \int _{0}^{t} f_{\ast }(s)\,ds$$ for all $$t\in I$$. Consider the linear continuous operator $$\theta \colon L^{1}(I,\mathbb{R})\to \mathcal{X}$$ defined by $$f\mapsto \theta (f)(t)$$, where $$\theta (f)(t)=a _{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds$$ for all $$t \in I$$. Since θ is a linear continuous map, by using Lemma 4 we get $$\theta \circ S_{\mathcal{F}}$$ is a closed graph operator. Note that $$g_{n} \in \theta \circ S_{\mathcal{F}}(x_{n})$$ for all n. Since $$x_{n} \to x_{\ast }$$ and $$g_{n} \to g_{\ast }$$, there exists $$f_{\ast }\in S_{\mathcal{F}}(x_{\ast })$$ such that $$g_{\ast }(t)=a _{\zeta }f_{\ast }(t)+b_{\zeta } \int _{0}^{t} f_{\ast }(s)\,ds$$ for all $$t\in I$$. For $$\lambda \in (0,1)$$ and $$x\in \lambda \varOmega (x)$$, there exists $$f\in S_{\mathcal{F},x}$$ such that $$x(t)= a_{\zeta }\lambda f(t)+b _{\zeta } \int _{0}^{t} \lambda f(s)\,ds$$ for all $$t\in I$$. Hence,

$$\bigl\vert x(t) \bigr\vert \leq \lambda (a_{\zeta }+b_{\zeta }) \eta ^{\ast } \cdot \Biggl(1+\zeta _{0}+\iota _{0}+ \sum _{i=1}^{m} \frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x \Vert .$$

Thus, $$\|x\|=\max_{t\in I}|x(t)|\leq \lambda \|x\|$$. Put $$\mathcal{W}= \{x\in \mathcal{X},\|x\|< r(1+\zeta _{0} +\iota _{0} + \sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}})\}$$. Note that the operator $$\varOmega \colon \overline{\mathcal{W}} \to P_{cp,cv}(\mathcal{X})$$ is upper semi-continuous and compact. In view of the choice of $$\mathcal{W}$$, there is no $$x\in \partial {\mathcal{W}}$$ such that $$x\in \lambda \varOmega (x)$$ for some $$\lambda \in (0,1)$$. Hence, by using Theorem 5, Ω has a fixed point $$x\in \overline{\mathcal{W}}$$ which is a solution for problem (1). This completes the proof. □

Now consider the Banach space $$\mathcal{X}=C(I)$$ via the norm

$$\Vert x \Vert =\max_{t\in I} \bigl\vert x(t) \bigr\vert + \sum_{i=1}^{m} \max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}x(t) \bigr\vert + \sum_{j=1}^{n} \max _{t\in I} \bigl\vert {} ^{\mathrm{CF}} \mathcal{I}^{\gamma _{j}}x(t) \bigr\vert .$$

Here, we investigate the fractional integro-differential inclusion

\begin{aligned}[b] {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta }x(t) \in{} & \mathcal{F} \bigl( t, x(t), (\phi x) (t),( \varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{2}} x(t),\ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{m}} x(t), \\ & {} {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t) \bigr), \end{aligned}
(2)

with boundary condition $$x(0)=0$$, where $$\zeta ,\beta _{1},\dots ,\beta _{m},\gamma _{1},\dots ,\gamma _{n}\in (0,1)$$. Similar to the last case, we say that a function $$x\in C(I,\mathbb{R})$$ is a solution for problem (2) whenever there exists a function $$f\in L^{1}(I)$$ such that

\begin{aligned} f(t) & \in \mathcal{F}\bigl(t,x(t),(\phi x) (t),( \varphi x) (t),{}^{\mathrm{CF}}_{ N}{ \mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \beta _{2}}x(t) , \ldots , \\ & \quad {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t) \bigr), ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{1}}x(t), ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t)) \end{aligned}

for almost all $$t\in I$$ and $$x(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0} ^{t} f(s)\,ds$$ for all $$t\in I$$.

### Theorem 12

Assume that $$\mathcal{F} \colon I \times \mathbb{R}^{m+n+3 }\to P_{cv,cp}( \mathbb{R})$$ is a multifunction such that the map $$t \to \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{3+m+n})$$ is measurable for all $$x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}$$, the map $$t \to d_{H}(0, \mathcal{F}(t,0,\ldots ,0))$$ is integrably bounded for almost all $$t\in I$$ and

\begin{aligned} & H_{d} \bigl( \mathcal{F}( t ,x_{1},x_{2},x_{3},y_{1},y_{2}, \ldots , y_{m},z_{1},z_{2},\ldots ,z_{n}), \\ & \qquad \mathcal{F}\bigl(t,x^{\prime }_{1},x^{\prime }_{2},x^{\prime } _{3},y^{\prime }_{1},y^{\prime }_{2}, \ldots ,y^{\prime }_{m},z^{ \prime }_{1},z^{\prime }_{2}, \ldots , z^{\prime }_{n}\bigr) \bigr) \\ &\quad \leq \eta (t) \Biggl( \bigl\vert x_{1}-x^{\prime }_{1} \bigr\vert + \bigl\vert x_{2}-x^{\prime }_{2} \bigr\vert + \bigl\vert x _{3}-x^{\prime }_{3} \bigr\vert +\sum_{i=1}^{m} \bigl\vert y_{i}-y^{\prime }_{i} \bigr\vert +\sum _{j=1} ^{n} \bigl\vert z_{j}-z^{\prime }_{j} \bigr\vert \Biggr) \end{aligned}

for all $$t\in I$$ and all $$x_{1},x_{2},x_{3},x^{\prime }_{1},x^{\prime }_{2},x^{\prime }_{3},y_{1},\ldots ,y_{m},y^{\prime }_{1},\ldots ,y ^{\prime }_{m},z_{1},\ldots ,z_{n},z^{\prime }_{1},\ldots ,z^{\prime }_{n} \in \mathbb{R}$$. If $$\Delta \leq 1$$, then the inclusion problem (2) has at least one solution, where

$$\Delta =\eta ^{\ast } \Biggl(1+n+\zeta _{0}+\iota _{0}+\sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Biggl(1+n+\sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr).$$

### Proof

By using the assumptions of Theorem III-6 in [34], we conclude that $$\mathcal{F}$$ admits a measurable selection $$f\colon I \to \mathbb{R}$$. Since $$\mathcal{F}$$ is integrable bounded, $$f\in L^{1}(I,\mathbb{R})$$ and so $$S_{\mathcal{F},x}$$ is nonempty for all $$x\in \mathcal{X}$$, where

\begin{aligned} S_{\mathcal{F},x} ={} & \bigl\{ f\in L^{1} (I,R): f(t) \in \mathcal{F}\bigl(t,x(t),( \phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1}} x(t), \\ & \quad {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t), {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , \\ & \quad {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t)\bigr) \text{ for all } t \in I\bigr\} . \end{aligned}

Define the operator $$\varOmega \colon \mathcal{X} \to P(\mathcal{X})$$ by

\begin{aligned} \varOmega (x)={}&\biggl\{ g\in \mathcal{X}: \mbox{ there exists } f\in S_{\mathcal{F},x} \mbox{ such that}\\ &g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds\mbox{ for all }t\in I\biggr\} . \end{aligned}

First, we show that $$\varOmega (x)\in P_{cl}(\mathcal{X})$$ for all $$x\in \mathcal{X}$$. Let $$g_{n}\in \varOmega (x)$$ for all $$n\geq 0$$ and $$g_{n}\rightarrow g_{\ast }$$ for some $$g\in \mathcal{X}$$. For each n, choose $$f_{n}\in S_{\mathcal{F},x}$$ such that $$g_{n}(t)= a_{\zeta }f _{n}(t)+b_{\zeta } \int _{0}^{t} f_{n}(s)\,ds$$ for all $$t\in I$$. Since $$\mathcal{F}$$ has compact values, there is a subsequence of $$f_{n}$$ that converges to f in $$L^{1}(I,\mathbb{R})$$. Thus, $$f\in S_{\mathcal{F},x}$$ and $$g_{n}(t)\rightarrow g_{\ast }(t)= a_{ \zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds$$ for all $$t\in I$$. This implies that $$g_{\ast }\in \varOmega$$. Now, we show that there exists $$\epsilon <1$$ such that $$H_{d}(\varOmega (x),\varOmega (y))\leq \epsilon \|x-y \|$$ for all $$x,y \in \mathcal{X}$$. Let $$x,y \in \mathcal{X}$$ and $$g_{1} \in \varOmega (x)$$. Choose $$f_{1} \in S_{\mathcal{F},x}$$ such that $$g_{1}(t)= a_{\zeta }f_{1}(t)+b_{\zeta } \int _{0}^{t} f_{1}(s)\,ds$$ for all $$t \in I$$. Consider the multifunction $$\tilde{\mathcal{F}}$$ defined by

\begin{aligned} \tilde{\mathcal{F}}\bigl(t,x(t)\bigr)={}&\mathcal{F}\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{2}} x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{m}} x(t), \\ &{}{}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots ,{}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t) \bigr). \end{aligned}

Then we have

\begin{aligned} H_{d}( \tilde{ \mathcal{F}}\bigl(t,x(t) \bigr), \tilde{\mathcal{F}}\bigl(t,y(t)\bigr)\leq{} & \eta (t) \Biggl( \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert (\phi x) (t)-(\phi y) (t) \bigr\vert \\ & {} + \bigl\vert (\varphi x) (t)-(\varphi y) (t) \bigr\vert \\ & {} + \sum_{i=1}^{m} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} x(t) - {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} y(t) \bigr\vert \\ & {} + \sum_{j=1}^{n} \bigl\vert ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}x(t)-^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}y(t) \bigr\vert \Biggr) \end{aligned}

for almost $$t\in I$$. Hence, there exists $$w_{t} \in \tilde{\mathcal{F}}(t,y(t))$$ such that

\begin{aligned} \bigl\vert f_{1}(t)-w_{t} \bigr\vert \leq {}& \eta (t) \Biggl( \bigl\vert x(t) - y(t) \bigr\vert + \bigl\vert (\phi x) (t)-( \phi y) (t) \bigr\vert + \bigl\vert (\varphi x) (t)-( \varphi y) (t) \bigr\vert \\ & {} + \sum_{i=1}^{m} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}} x(t) - {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} y(t) \bigr\vert \\ & {} + \sum_{j=1}^{n} \bigl\vert ^{\mathrm{CF}}\mathcal{I}^{ \gamma _{j}}x(t)-^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}y(t) \bigr\vert \Biggr):=M_{t} \end{aligned}

for almost $$t\in I$$. Define $$V\colon I \to P(\mathbb{R})$$ by $$V(t)=\{u\in \mathbb{R}: |f_{1}(t)-u|\leq M_{t}\}$$ for all $$t\in I$$. By using Theorem III-41 in [34], we get V is measurable. Since $$t \mapsto V(t)\cap \tilde{\mathcal{F}}(t,y(t))$$ is measurable (Proposition III-4 in [34]), we can choose $$f_{2} \in S_{\mathcal{F},y}$$ such that $$|f_{1}(t)-f_{2}(t)|\leq M _{t}$$ for almost all $$t\in I$$. Define $$g_{2} \in \varOmega (y)$$ by $$g_{2}(t)= a_{\zeta }f_{2}(t)+b_{\zeta } \int _{0}^{t} f_{2}(s)\,ds$$ for all $$t \in I$$. Then we have

\begin{aligned} \Vert g_{1}-g_{2} \Vert ={}&\max_{t\in I} \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert +\sum _{i=1}^{m} \max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}g_{1}(t)-{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}g_{2}(t) \bigr\vert \\ & {} + \sum_{i=1}^{n} \max _{t\in I} \bigl\vert {}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g _{1}(t)-{}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g_{2}(t) \bigr\vert \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ \leq {}& a_{\zeta } \bigl\vert f_{1}(t)-f_{2}(t) \bigr\vert +b_{\zeta } \int _{0}^{t} \bigl\vert f_{1}(s)-f _{2}(s) \bigr\vert \,ds \\ \leq {}&\eta (t) \Biggl(1+n+\zeta _{0}+\iota _{0}+\sum _{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) (a_{\zeta }+ b_{\zeta }) \Vert x-y \Vert , \end{aligned}

and so

\begin{aligned} &\bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}g_{1}(t)-{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{i}}g_{2}(t) \bigr\vert \\ &\quad \leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ &\quad \leq \eta (t)\frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggl(1+n+\zeta _{0}+\iota _{0} + \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) (a _{\zeta }+ b_{\zeta }) \Vert x-y \Vert . \end{aligned}

Thus,

\begin{aligned} \bigl\vert {}^{\mathrm{CF}} \mathcal{I}^{\gamma _{i}}g_{1}(t) - {}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g_{2}(t) \bigr\vert & \leq \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ & \leq \eta (t) \Biggl(1 + n+\zeta _{0} + \iota _{0} +\sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) ( a _{\zeta }+ b_{\zeta }) \Vert x-y \Vert , \end{aligned}

and so

\begin{aligned} \Vert g_{1}-g_{2} \Vert \leq{}& \eta ^{\ast } \Biggl(1 + n+ \zeta _{0}+\iota _{0} + \sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \\ & {} \times \Biggl(1+n +\sum_{i=1}^{m} \frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert = \Delta \Vert x-y \Vert . \end{aligned}

Hence, $$H_{d}(\varOmega (x), \varOmega (y))\leq \Delta \|x-y\|$$. Since $$\Delta <1$$, Ω is a closed-valued contraction. By using Theorem 6, Ω has a fixed point which is a solution for the inclusion problem (2). □

Consider the Banach space $${\mathcal{X}}=\{x: x,{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}}x \in C(I,\mathbb{R})\}$$ endowed with the norm $$\|x\|= \max_{t\in I} |x(t)|+\max_{t\in I} |{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{i}}x(t)|$$. Here, we review the inclusion problem

\begin{aligned}[b] {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta }x(t) \in{}& \mathcal{F}\bigl(t,x(t),( \phi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}} _{N}{\mathcal{D}}^{\beta _{2}} x(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{n}} x(t)\bigr) \\ & {} +\mathcal{G}\bigl(t,x(t),(\varphi x) (t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}} \mathcal{I}^{ \beta _{n}}x(t)\bigr) \end{aligned}
(3)

with boundary condition $$x(0)=0$$, where $$\zeta ,\beta _{1},\dots ,\beta _{n}\in (0,1)$$. Define the set of the selections of $$\mathcal{F}$$ and $$\mathcal{G}$$ at x by

\begin{aligned} S_{\mathcal{F},x}={}&\bigl\{ v\in L^{1}[0,1]:v(t)\in \mathcal{F} \bigl(t,x(t),( \phi x) (t), \\ &{}{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{2}} x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t)\bigr)\text{ for almost all } t\in I\bigr\} \end{aligned}

and

\begin{aligned} S_{\mathcal{G},x}={}&\bigl\{ v\in L^{1}[0,1]:v(t)\in \mathcal{G} \bigl(t,x(t),( \varphi x) (t), \\ &{}{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{2}}x(t), \ldots ,{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t) \bigr)\text{ for almost all }t \in I\bigr\} . \end{aligned}

We suppose that $$S_{\mathcal{F},x}\neq \emptyset$$ and $$S_{ \mathcal{G},x}\neq \emptyset$$ for all $$x\in {\mathcal{X}}$$. A function $$x\in C(I,\mathbb{R})$$ is a solution for problem (3) whenever there exist two functions $$f\in H^{1}(I)$$ and $$f^{\prime }\in H^{1}(I)$$ such that

$$f(t) \in \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{n}}x(t) \bigr)$$

and $$f^{\prime } \in \mathcal{G}(t,x(t),(\varphi x)(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t))$$ for almost all $$t\in I$$ and

$$x(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds + a_{\zeta }f^{ \prime }(t)+b_{\zeta } \int _{0}^{t} f^{\prime }(s)\,ds$$

for all $$t\in I$$.

### Theorem 13

Let $$\mathcal{F}: I\times \mathbb{R}^{n+2}\to P_{cp,cv}(\mathbb{R})$$ be a multifunction and $$\mathcal{G}: I\times \mathbb{R}^{n+2}\to P_{cp,cv}( \mathbb{R})$$ be a Caratheodory set-valued map. Assume that there exist continuous functions $$p, m:I\to (0,\infty )$$ and $$\eta (t) \in L^{ \infty }(I)$$ such that $$t\vdash \mathcal{F}(t,y_{1},\ldots ,y_{n+2})$$ is measurable,

\begin{aligned}& \bigl\Vert \mathcal{F} \bigl(t,x(t),(\phi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{n}}x(t) \bigr) \bigr\Vert \leq m(t), \\& \bigl\Vert \mathcal{G}\bigl(t,x(t),(\varphi x) (t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t)\bigr) \bigr\Vert \leq p(t), \end{aligned}

and

\begin{aligned}& H_{d}\bigl( \mathcal{F}(t,y_{1}, \ldots , y_{n+2}),\mathcal{F}\bigl(t,y^{\prime }_{1}, \ldots , y^{\prime }_{n+2}\bigr)\bigr) \leq \eta (t)\sum _{i=1}^{n+2} \bigl( \bigl\vert y _{i}-y^{\prime }_{i} \bigr\vert \bigr) \end{aligned}

for all $$t\in I$$, $$x\in {\mathcal{X}}$$ and $$y_{1}, \ldots , y_{n+2}$$, $$y'_{1}, \ldots , y'_{n+2}\in \mathbb{R}$$. If $$L=\eta ^{\ast }(1+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})(1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})< 1$$, then the inclusion problem (3) has at least one solution.

### Proof

Put $$\mathcal{Y}=\{x\in \mathcal{X}: \|x\|\leq M\}$$, where $$M=(1+ \sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})(\|p\|_{\infty }+ \|m\|_{\infty })$$. One can check that $$\mathcal{Y}$$ is a closed, bounded, and convex subset of $$\mathcal{X}$$. Define the multivalued operators $$\mathcal{A},\mathcal{B}: \mathcal{Y}\to P(\mathcal{X})$$ by

$$\mathcal{A}x:=\biggl\{ x\in {\mathcal{X}}: ~\text{there is}~v\in S_{ \mathcal{F},x}~\text{such that}~x(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds~\text{for all}~t\in I\biggr\}$$

and $$\mathcal{B}x:=\{x\in {\mathcal{X}}:~\text{there is}~v\in S_{ \mathcal{G},x}~\text{such that}~x(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds~\text{for all}~t\in I\}$$. Note that problem (3) is equivalent to the inclusion fixed point problem $$x\in \mathcal{A}x+\mathcal{B}x$$. Also, the operator $$\mathcal{A}$$ is equivalent to the composition $$\theta \circ S_{\mathcal{F}}$$, where θ is the continuous linear operator on $$L^{1}(0,1)$$ into $$\mathcal{X}$$ defined by $$\theta v(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds$$. Let $$x\in \mathcal{Y}$$ and $$\{v_{n}\}_{n\geq 1}$$ be a sequence in $$S_{\mathcal{F},x}$$. Then $$v_{n}(t)\in \mathcal{F}(t,x(t),( \phi x)(t),{}^{\mathrm{CF}}{\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}{\mathcal{D}} ^{ \beta _{2}}x(t) ,\ldots ,{}^{\mathrm{CF}}{\mathcal{D}}^{ \beta _{n}}x(t) )$$ for almost $$t\in I$$. Since

$$\mathcal{F}\bigl(t,x(t),(\phi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1} } x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}} x(t) ,\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{n}}x(t) \bigr)$$

is compact for all $$t\in I$$, there is a convergent subsequence of $$\{v_{n}(t)\}$$ (call it again $$\{v_{n}(t)\}$$) such that it converges in measure to some $$v(t)\in S_{\mathcal{F},x}$$ for almost all $$t\in I$$. Since θ is continuous, $$\theta v_{n}(t)\to \theta v(t)$$ pointwise on I. In order to show that the convergence is uniform, we show that $$\{\theta v_{n}\}$$ is an equi-continuous sequence. Let $$\tau < t \in I$$. Then we have

$$\bigl\vert \theta v_{n}(t)- \theta v_{n}(\tau ) \bigr\vert \leq a_{\zeta } \bigl\vert v_{n}(t)-v_{n}( \tau ) \bigr\vert +b_{\zeta } \int _{\tau }^{t} \bigl\vert v_{n}(s) \bigr\vert \,ds.$$

Since the right-hand of the above inequality tends to 0 as $$t\to \tau$$, the sequence $$\{\theta v_{n}\}$$ is equi-continuous. Now, by using the Arzela–Ascoli theorem, there is a uniformly convergent subsequence of $$\{v_{n}\}$$ (we show it again by $$\{v_{n}\}$$) such that $$\theta v_{n}\to \theta v$$. Note that $$\theta v \in \theta (S_{ \mathcal{F},x})$$. Hence, $$\mathcal{A}x=\theta (S_{\mathcal{F},x})$$ is compact for all $$x\in \mathcal{Y}$$. Now, we show that $$\mathcal{A}x$$ is convex for all $$x\in \mathcal{Y}$$. Let $$u , u^{\prime }\in \mathcal{A}x$$. Choose $$v,v^{\prime }\in S_{\mathcal{F},x}$$ such that $$u(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds$$ and $$u'(t)=a_{ \zeta }v^{\prime }(t)+b_{\zeta } \int _{0}^{t}v^{\prime }(s)\,ds$$ for almost all $$t\in I$$. Let $$0\leq \lambda \leq 1$$. Then we have

$$\bigl(\lambda u + (1-\lambda ) u^{\prime } \bigr) (t) = a_{\zeta } \bigl(\lambda v(t)+(1-\lambda )v^{\prime }(t) \bigr) +b_{\zeta } \int _{0}^{t} \bigl(\lambda v(s)+(1-\lambda )v^{ \prime }(s) \bigr)\,ds.$$

Since $$\mathcal{F}$$ is convex-valued, $$\lambda u+(1-\lambda )u^{ \prime }\in \mathcal{A}x$$. Similarly, we can show that $$\mathcal{B}$$ is compact and convex-valued. Here, we show that $$\mathcal{A}y+ \mathcal{B}y\subset \mathcal{Y}$$ for all $$y\in \mathcal{Y}$$. Let $$y\in \mathcal{Y}$$, $$u\in \mathcal{A}y$$, and $$u^{\prime }\in \mathcal{B}y$$. Choose $$v\in S_{\mathcal{F},y}$$ and $$v'\in S_{ \mathcal{G},y}$$ such that $$u(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0} ^{t}v(s)\,ds$$ and $$u'(t)=a_{\zeta }v^{\prime }(t)+b_{\zeta } \int _{0} ^{t}v^{\prime }(s)\,ds$$ for almost all $$t\in I$$. Hence,

$$\bigl\vert u(t)+u'(t) \bigr\vert \leq a_{\zeta } \bigl( \bigl\vert v(t) \bigr\vert + \bigl\vert v^{\prime }(t) \bigr\vert \bigr)+b_{\zeta } \int _{0}^{t} \bigl( \bigl\vert v(s) \bigr\vert + \bigl\vert v^{\prime }(s) \bigr\vert \bigr)\,ds,$$

and so

\begin{aligned} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)+{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}}u'(t) \bigr\vert \leq{} & \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u(t) \bigr\vert + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \\ \leq {}&\frac{ a_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}}\bigl(p(t)+m(t)\bigr) \\ & {} + \frac{ b_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \end{aligned}

for $$1\leq i \leq n$$. This implies that

$$\max_{t\in I} \bigl\vert u(t)+u'(t) \bigr\vert \leq a_{\zeta } \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{ \infty }\bigr)+b_{\zeta } \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr)= \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }$$

and

\begin{aligned} \max_{t \in I} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)+{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \leq {}&\frac{ a_{\zeta }B( \beta _{i})}{(1-\beta _{i})^{2}}\bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \\ & {} + \frac{ b_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \\ = {}&\frac{ B(\beta _{i})( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty })}{(1-\beta _{i})^{2}}. \end{aligned}

Thus, $$\|u+u^{\prime }\|\leq (1+\sum_{i=1}^{n}(\frac{B(\beta _{i})}{(1- \beta _{i})^{2}})) (\|p\|_{\infty }+\|m\|_{\infty })=M$$. Now, we show that the operator $$\mathcal{B}$$ is compact on $$\mathcal{Y}$$. To do this, we prove that $$\mathcal{B}(\mathcal{Y})$$ is uniformly bounded and equi-continuous in $$\mathcal{X}$$. Let $$u\in \mathcal{B}(\mathcal{Y})$$ be arbitrary. Choose $$v\in S_{\mathcal{G},x}$$ such that $$u(t)=a_{ \zeta }v(t)+b_{\zeta }\int _{0}^{t} v(s)\,ds$$ for some $$x\in \mathcal{Y}$$. Hence,

\begin{aligned} \bigl\vert u(t) \bigr\vert & \leq a_{\zeta } \bigl\vert v(t) \bigr\vert +b_{\zeta } \int _{0}^{t} \bigl\vert v(s) \bigr\vert \,ds \bigl|{} ^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{i}}u(t))\bigr| \\ & \leq a_{\zeta } \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}v(t) \bigr\vert +b _{\zeta } \int _{0}^{t} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}v(s) \bigr\vert \,ds \\ & \leq \frac{B(\beta _{i})(a_{\zeta }+ b_{\zeta })}{(1-\beta _{i})^{2}} p(t) \\ & =\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} p(t). \end{aligned}

Thus, $$\max_{t\in I}|u(t)|\leq (a_{\zeta }+b_{\zeta })\|p\|_{\infty }= \|p\|_{\infty }$$ and $$\max_{t\in I}|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{i}}u_{i}(t)|\leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}\|p\| _{\infty }$$ for $$i=1,\dots ,n$$, and so $$\|u\| \leq (1+\sum_{i=1}^{n} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}})\|p\|_{\infty }$$. Here, we show that $$\mathcal{B}$$ maps $$\mathcal{Y}$$ to equi-continuous subsets of $$\mathcal{X}$$. Let $$t, \tau \in I$$ with $$\tau < t$$, $$x\in \mathcal{Y}$$ and $$u \in \mathcal{B}x$$. Choose $$v\in S_{\mathcal{G},x}$$ such that $$u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t} v(s)\,ds$$ for all. Then we have

$$\bigl\vert u(t)- u(\tau ) \bigr\vert \leq a_{\zeta }\bigl(v(t)-v(\tau ) \bigr)+b_{\zeta } \int _{\tau } ^{t}v(s)\,ds \leq a_{\zeta } \bigl(v(t)-v(\tau )\bigr)+b_{\zeta }(t-\tau ) \Vert p \Vert _{\infty }$$

and $$|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u(t)- {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{i}}u(\tau )|\leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}|u(t)-u(\tau )|$$. Since the right-hand of the inequality tends to 0 as $$t\to \tau$$, by using the Arzela–Ascoli theorem, we get $$\mathcal{B}$$ is compact. Now, we show that $$\mathcal{B}$$ has a closed graph. Let $$x_{n}\in \mathcal{Y}$$ and $$u_{n}\in \mathcal{B}(x_{n})$$ for all n with $$x_{n}\to x_{0}$$ and $$u_{n}\to u_{0}$$. We show that $$u_{0}\in \mathcal{B}(x_{0})$$. For each n, choose $$v_{n}\in S_{ \mathcal{G},x_{n}}$$ such that $$u_{n}(t)= a_{\zeta }v_{n}(t)+b_{ \zeta }\int _{0}^{t}v_{n}(s) \,ds$$ for all $$t\in I$$. Again, consider the continuous linear operator $$\theta :L^{1}(0,1)\to {\mathcal{X}}$$ defined by $$\theta (v)(t)= a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s) \,ds$$. By using Lemma 4, $$\theta o S_{\mathcal{G}}$$ is a closed graph operator. Since $$u_{n}\in \theta (S_{\mathcal{G},x_{n}})$$ for all n and $$x_{n}\to x_{0}$$, there exists $$v_{0}\in S_{\mathcal{G},x_{0}}$$ such that $$u_{0}(t)= a_{\zeta }v_{0}(t)+b_{\zeta }\int _{0}^{t}v_{0}(s) \,ds$$. Hence, $$u_{0}\in \mathcal{B}(x_{0})$$. This implies that $$\mathcal{B}$$ has a closed graph, and so $$\mathcal{B}$$ is upper semi-continuous. Now, we show that $$\mathcal{A}$$ is a contraction multifunction. Let $$x,y\in {\mathcal{X}}$$ and $$u\in \mathcal{A}y$$. Choose $$v\in S_{\mathcal{F},y}$$ such that $$u(t)= a_{\zeta }v(t)+b_{ \zeta }\int _{0}^{t}v(s) \,ds$$ for all $$t\in I$$. Since

\begin{aligned} &H_{d} \bigl( \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{1}}x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t)\bigr), \\ & \qquad \mathcal{F}\bigl(t,y(t),(\phi y) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{1}}y(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{n}} y(t)\bigr) \bigr) \\ & \quad \leq \eta (t) \Biggl(1 + \zeta _{0} + \sum _{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert \end{aligned}

for almost all $$t\in I$$, there exists $$w\in \mathcal{F}(t,x(t),( \phi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t))$$ such that $$|v(t)-w| \leq \eta (t) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} )\|x-y\|$$ for almost all $$t\in I$$. Consider the multifunction $$U:I\to 2^{\mathbb{R}}$$ defined by

$$U(t)=\Biggl\{ w\in \mathbb{R}: \bigl\vert v(t)-w \bigr\vert \leq \eta (t) \Biggl(1+\zeta _{0}+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \text{ for almost all } t\in I \Biggr\} .$$

Since v and $$\eta (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} )$$ are measurable, we get

$$U(\cdot)\cap \mathcal{F}\bigl(t,x(\cdot),(\phi x) (\cdot),{}^{\mathrm{CF}}{ \mathcal{D}}^{\beta _{1}}x(\cdot), \ldots ,{}^{\mathrm{CF}}{ \mathcal{D}}^{\beta _{n}} x(\cdot)\bigr)$$

is a measurable multifunction. Choose

$$v^{\prime }(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t), \ldots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{n}} x(t) \bigr)$$

such that $$|v(t)-v'(t)|\leq \eta (t) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B( \beta _{i})}{(1-\beta _{i})^{2}} )\|x-y\|$$ and $$u^{\prime }(t)=a_{ \zeta }v^{\prime }(t)+b_{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$$ for all $$t\in I$$. Since $$|u(t)-u'(t)|\leq a_{\zeta }(v(t)- v^{\prime }(t))+b _{\zeta }\int _{0}^{t}(v(s)-v^{\prime }(s))\,ds$$ and

$$\bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}} u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}} u^{\prime }(t) \bigr\vert \leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl\vert u(t)-u^{\prime }(t) \bigr\vert ,$$

we get

\begin{aligned} \max_{t\in I} \bigl\vert u(t)-u'(t) \bigr\vert \leq{}& a_{\zeta }\eta ^{\ast } \Biggl(1 + \zeta _{0}+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \\ & {}+b_{\zeta }\eta ^{\ast } \Biggl(1 +\zeta _{0} + \sum_{i=1}^{n}\frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \\ ={}&\eta ^{\ast } \Biggl(1+\zeta _{0} + \sum _{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert \end{aligned}

and

$$\max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)-{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \leq \eta ^{\ast }\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggl(1+\zeta _{0}+\sum _{i=1}^{n}\frac{1}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert |$$

for $$1\leq i\leq n$$. Hence, $$\|u-u^{\prime }\| \leq \eta ^{\ast } (1+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} ) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )\|x-y\|$$. This implies that $$H_{d}(\mathcal{A}x, \mathcal{A}y) \leq L \|x-y\|$$. Now, by using Theorem 7, the inclusion fixed point problem $$x\in \mathcal{A}x +\mathcal{B}x$$ has a solution which is a solution for the inclusion problem (3). □

Now, we are ready to investigate the fractional integro-differential inclusion

\begin{aligned}& {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta } \biggl( \frac{x(t)}{g(t,x(t),( \phi x)(t),(\varphi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{1}}x(t), \dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t))} \biggr) \\& \quad \in \mathcal{G}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{k}}x(t)\bigr) \end{aligned}
(4)

with boundary condition $$u(0)=0$$, where $$\zeta ,\zeta _{1},\dots ,\zeta _{n},\beta _{1},\dots ,\beta _{k}\in (0,1)$$, $$g: I\times \mathbb{R}^{n+3} \to \mathbb{R}\backslash \{0\}$$ is continuous and $$\mathcal{G}:I \times \mathbb{R}^{k+3}\to \mathcal{P}(\mathbb{R})$$ is a multifunction. We say that $$x\in \mathcal{X}$$ is a solution for problem (4) whenever it satisfies the boundary conditions and there exists $$v\in S_{ \mathcal{G},x}$$ such that

\begin{aligned} x(t) ={}& g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta _{1}}x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & {} \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t} v(s)\,ds \biggr), \end{aligned}

where

\begin{aligned} S_{\mathcal{G},x} ={}& \bigl\{ v\in L^{1}[0,1]:v(t) \in \mathcal{G}\bigl(t,x(t),( \phi x) (t),(\varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{k}}x(t) \bigr) \text{ for almost all } t \in I \bigr\} . \end{aligned}

### Theorem 14

Suppose that $$\mathcal{G}: I\times \mathbb{R}^{k+3}\to \mathcal{P} _{cp,cv}(\mathbb{R})$$ is a Caratheodory set-valued map, $$g: J\times \mathbb{R}^{n+3}\to \mathbb{R}\backslash \{0\}$$ is a bounded continuous map with upper bound K and there are continuous functions $$p, m:J\to (0,\infty )$$ such that $$\|\mathcal{G}(t, x_{1}, x_{2}, \ldots ,x_{k+3})\|\leq m(s)$$ and

$$\bigl\vert g(t,x_{1}, x_{2},\ldots ,x_{n+3})-g(t,y_{1}, y_{2},\ldots ,y_{n+3}) \bigr\vert \leq \eta (t)\sum _{i=1}^{n+3} \vert x_{i}-y_{i} \vert$$

for all $$t\in I$$. If $$\eta ^{\ast }(1+\zeta _{0}+\iota _{0}+ \sum_{i=1} ^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}})\cdot K\cdot \|m\|_{\infty }<1$$, then the inclusion problem (4) has a solution.

### Proof

Put $$S=\{x\in {\mathcal{X}}: \|x\|\leq L\}$$, where $$L=K\|m\|_{\infty }$$. It is clear that S is a convex, closed, and bounded subset of the Banach space $$\mathcal{X}$$. Define $$\mathcal{A}, \mathcal{B} : S \to \mathcal{P}({\mathcal{X}})$$ by

$$\mathcal{A}x(t)=g\bigl\{ t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N} { \mathcal{D}}^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{ \zeta _{n}}x(t)\bigr\}$$

and

$$\mathcal{B}x(t)=\biggl\{ u\in {\mathcal{X}}: \text{ there is } v\in S_{ \mathcal{G},x} \text{ such that } u(t)= a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \text{ for all } t\in I\biggr\} .$$

Thus, the problem of fractional differential inclusions is equivalent to the inclusion problem $$x\in \mathcal{A}(x)\mathcal{B}(x)$$. Consider the operator $$\mathcal{B}=\theta \circ S_{\mathcal{G}}$$, where θ is the continuous linear operator on $$L^{1}(I)$$ into $$\mathcal{X}$$ defined by $$\theta v(s)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s)\,ds$$. Let $$x\in S$$ be arbitrary and $$\{v_{n}\}$$ be a sequence in $$S_{\mathcal{G},x}$$. Then $$v_{n}(t)\in \mathcal{G}(t,x(t),(\phi x)(t),( \varphi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{k}}x(t) )$$ for almost $$t\in I$$. Since

$$\mathcal{G}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N} { \mathcal{D}}^{\beta _{k}}x(t) \bigr)$$

is compact for all $$t\in I$$, there is a convergent subsequence of $$\{v_{n}(t)\}$$ (show it by $$\{v_{n}(t)\}$$ again) to some $$v\in S_{ \mathcal{G},x}$$. Note that $$\theta v_{n}(t)\to \theta v(t)$$ pointwise on I because θ is continuous. Now, we show that $$\{\theta v _{n}\}$$ is an equi-continuous sequence. Let $$\tau < t\in I$$. Then we have $$|\theta v_{n}(t)- \theta v_{n}(\tau )|\leq a_{\zeta }|v_{n}(t)-v _{n}(\tau )|+b_{\zeta }\int _{\tau }^{t} |v_{n}(s)| \,ds$$. Thus, the sequence $$\{\theta v_{n}\}$$ is equi-continuous because the right-hand of the inequality tends to 0 as $$t\to \tau$$. Hence, it has a uniformly convergent subsequence by using the Arzela–Ascoli theorem. Choose a subsequence of $$\{v_{n}\}$$ (we show it again by $$\{v_{n}\}$$) such that $$\theta v_{n}\to \theta v$$. Hence, $$\theta v \in \theta (S _{\mathcal{G} ,x})$$ and so $$\mathcal{B}=\theta (S_{\mathcal{G} ,x})$$ is compact for all $$x\in S$$. Here, we prove that $$\mathcal{B}x$$ is convex for all $$x\in S$$. Let $$x\in S$$ and $$u, u^{\prime }\in \mathcal{B}x$$. Choose $$v,v^{\prime }\in S_{\mathcal{G},x}$$ such that $$u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s)\,ds$$ and $$u'(t)=a_{\zeta }v^{\prime }(t)+b _{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$$ for almost all $$t\in I$$. Let $$0\leq \lambda \leq 1$$. Then we have

$$\lambda u(t)+(1-\lambda )u^{\prime }(t)=a_{\zeta } \bigl(\lambda v(t)+(1- \lambda ) v^{\prime }(t) \bigr)+b_{\zeta } \int _{0}^{t} \bigl(\lambda v(s)+(1- \lambda ) v^{\prime }(s) \bigr)\,ds.$$

Since $$\mathcal{G}$$ is convex-valued, $$\lambda u+(1-\lambda )u^{ \prime }\in \mathcal{B}x$$. It is clear that $$\mathcal{A}$$ is bounded, closed, and convex-valued. We show that $$\mathcal{A}x \mathcal{B}x$$ is a convex subset of S for all $$x\in S$$. Let $$x\in S$$ and $$u,u'\in \mathcal{A}x \mathcal{B}x$$. Choose $$v,v'\in S_{\mathcal{G},x}$$ such that

$$u(t)=g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr),$$

and

\begin{aligned} u'(t) & =g\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \zeta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & \quad \times \biggl(a_{\zeta }v^{\prime }(t)+b_{\zeta } \int _{0}^{t}v ^{\prime }(s)\,ds \biggr) \end{aligned}

for almost all $$t\in I$$. Hence,

\begin{aligned} \lambda u(t) + (1-\lambda )u^{\prime }(t) ={}& g \bigl(t,x(t),(\phi x) (t),( \varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & {}\times \biggl[a_{\zeta }\bigl(\lambda v(t)+(1-\lambda )v^{\prime }(t)\bigr) \\ & {} + b_{\zeta } \int _{0}^{t}\bigl(\lambda v(s)+(1-\lambda )v^{\prime }(s)\bigr)\,ds \biggr]. \end{aligned}

Note that $$\lambda u+(1-\lambda )u^{\prime }\in \mathcal{A}x \mathcal{B}x$$ because $$\mathcal{G}$$ is convex-valued. Hence, $$\mathcal{A}x\mathcal{B}x$$ is a convex subset of $$\mathcal{X}$$ for all $$x\in \mathcal{X}$$. However, we have

\begin{aligned} \bigl\vert u(t) \bigr\vert &= \biggl\vert g\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\zeta _{1}}x(t), \dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr) \biggr\vert \\ &\leq K(a_{\zeta }+b_{\zeta }) \Vert m \Vert _{\infty }=L< 1 \end{aligned}

for all $$t\in I$$, and so $$u\in S$$ and $$\mathcal{A}x\mathcal{B}x$$ is a convex subset of S for all $$x\in S$$. Now, we show that the operator $${\mathcal{B}}$$ is compact. It is enough to prove that $$\mathcal{B}(S)$$ is uniformly bounded and equi-continuous. Let $$u\in \mathcal{B}(S)$$. Choose $$v\in S_{\mathcal{G},x}$$ such that

$$u(t)=g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr)$$

for some $$x\in S$$. Since $$|u(t)|\leq K(a_{\zeta }+b_{\zeta })\|m\| _{\infty }$$, $$\|u\|_{\infty }=\max_{t\in I}|u(t)|\leq K(a_{\zeta }+b _{\zeta })\|m\|_{\infty }$$. Now, we prove that $$\mathcal{B}$$ maps S to equi-continuous subsets of $$\mathcal{X}$$. Let $$t, \tau \in J$$ with $$\tau < t$$, $$x\in S$$, and $$u \in \mathcal{B}x$$. Choose $$v\in S _{\mathcal{G},x}$$ such that $$u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0} ^{t}v(s)\,ds$$. Then we have

$$\bigl\vert u(t)- u(\tau ) \bigr\vert \leq a_{\zeta } \bigl\vert v(t)-v(\tau ) \bigr\vert +b_{\zeta } \int _{\tau } ^{t} \bigl\vert v(s) \bigr\vert \,ds.$$

Note that the right-hand side of this inequality tends to 0 as $$t\to \tau$$. By using the Arzela–Ascoli theorem, we get $$\mathcal{B}$$ is compact. Here, we show that $$\mathcal{B}$$ has a closed graph. Let $$x_{n}\in S$$ and $$u_{n}\in \mathcal{B}x_{n}$$ for all n with $$x_{n}\to x'$$ and $$u_{n}\to u'$$. We show that $$u'\in \mathcal{B}x'$$. For each n, choose $$v_{n}\in S_{\mathcal{G},x_{n}}$$ such that $$u_{n}(t)=a_{\zeta }v_{n}(t)+b_{\zeta }\int _{0}^{t}v_{n}(s)\,ds$$ for all $$t\in J$$. Again, consider the continuous linear operator $$\theta :L ^{1}(I)\to {\mathcal{X}}$$ such that $$\theta (v)(t)=u(t)=a_{\zeta }v(t)+b _{\zeta }\int _{0}^{t}v(s)\,ds$$. By using Lemma 4, $$\theta \circ S _{\mathcal{G}}$$ is a closed graph operator. Since $$x_{n}\to x'$$ and $$u_{n}\in \theta (S_{\mathcal{G},x_{n}})$$ for all n, there is $$v'\in S_{\mathcal{G},x'}$$ such that $$u'(s)=a_{\zeta }v^{\prime }(t)+b _{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$$. Hence, $$u'\in \mathcal{B}x'$$. Thus, $$\mathcal{B}$$ has a closed graph and so $$\mathcal{B}$$ is upper semi-continuous. Finally note that

\begin{aligned} H(\mathcal{A} x, \mathcal{A}y) =& \Vert \mathcal{A}x-\mathcal{A}y \Vert \\ =&\max_{t\in I}\bigl| g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta _{1}}x(t),\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta _{n}}x(t)\bigr) \\ &{} -g\bigl(t,y(t),(\phi y) (t),(\varphi y) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}y(t),\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}y(t)\bigr))\bigr| \\ \leq& \max_{t\in I} \bigl\vert \eta (t) \bigr\vert \Biggl( 1 + \zeta _{0}+\iota _{0}+\sum_{i=1}^{n} \frac{B(\zeta _{i})}{(1-\zeta _{i})^{2}} \Biggr) \bigl\vert x(t)-y(t) \bigr\vert \\ =&\eta ^{\ast } ( \Biggl(1+\zeta _{0}+\iota _{0}+ \sum_{i=1}^{n}\frac{B( \zeta _{i})}{(1-\zeta _{i})^{2}} \Biggr) \Vert x-y \Vert _{\infty } \end{aligned}

for all $$x,y\in {\mathcal{X}}$$. Now, by using Theorem 8, the inclusion problem $$x\in \mathcal{A}x \mathcal{B}x$$ has a solution which is a solution for problem (4). □

In this part, we show that the set of solutions for the second fractional integro-differential inclusion problem is infinite dimensional under some conditions. First we prove the next result.

### Lemma 15

Suppose that $$m\in L^{1}(I,\mathbb{R}^{+})$$, $$\mathcal{F}: I\times \mathbb{R}^{m+n+3}\to \mathcal{P}_{cv,cp}(\mathbb{R})$$ is a multivalued map such that the map $$t\vdash f(t,x_{1},x_{2},\ldots ,x_{3+m+n})$$ is measurable and

$$\bigl\Vert \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3}) \bigr\Vert =\sup \bigl\{ \vert f \vert : f\in \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\bigr\} \leq m(t)$$

for almost all $$t\in I$$ and $$\in x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}$$. Define $$\varPhi :{\mathcal{X}}\to \mathcal{P}({\mathcal{X}})$$ by

$$\varPhi (x)= \biggl\{ g\in {\mathcal{X}}: \textit{there is } f\in S_{ \mathcal{F},x} \textit{ such that } g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0} ^{t}f(s)\,ds \textit{ for all } t\in I \biggr\} .$$

Then $$\varPhi (x)\in \mathcal{P}_{cp.cv}({\mathcal{X}})$$ for all $$x\in {\mathcal{X}}$$.

### Proof

Note that $$\varPhi =\theta \circ S_{\mathcal{F}}$$, where $$\theta :L^{1}(I, \mathbb{R})\to \mathcal{X}$$ is the continuous linear map defined by $$\theta g(t)=a_{\zeta } f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds$$. Let $$x\in {\mathcal{X}}$$ and $$\{g_{n}\}$$ be a sequence in $$S_{\mathcal{F},x}$$. Then we have

$$g_{n}(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{1}}x(t),\dots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t), {}^{cF}\mathcal{I}^{\gamma _{1}}x(t),\dots ,{}^{cF} \mathcal{I}^{\gamma _{n}}x(t)\bigr)$$

for almost $$t\in I$$. Since

$$\mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t),\dots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{m}}x(t), {}^{cF} \mathcal{I}^{\gamma _{1}}x(t), \dots ,{}^{cF}\mathcal{I}^{\gamma _{n}}x(t)\bigr)$$

is compact for all $$t\in I$$, there is a convergent subsequence of $$\{g_{n}(t)\}$$ (show it by $$\{g_{n}(t)\}$$) which converges to some $$g\in S_{\mathcal{F},x}$$. Note that $$\theta g_{n}(t)\to \theta g(t)$$ pointwise on I because θ is continuous. Here, we prove that $$\{\theta g_{n}\}$$ is an equi-continuous sequence. Let $$\tau < t \in I$$. Then we have $$|\theta g_{n}(t)- \theta g_{n}(\tau )|= a_{ \zeta } (f(t)-f(\tau ) )+b_{\zeta } \int _{\tau }^{t}f(s)\,ds$$. Note that the sequence $$\{\theta g_{n}\}$$ is equi-continuous because the right-hand side of the inequality tends to zero when $$\tau \to t$$. Thus, there is a uniformly convergent subsequence of $$\{g_{n}\}$$ (show it by $$\{g_{n}\}$$ again) such that $$\theta g_{n}\to \theta g$$ (we use the Arzela–Ascoli theorem). This implies that $$\theta g \in \theta (S _{\mathcal{F},x})$$. Hence, $$\varPhi x=\theta (S_{\mathcal{F},x})$$ is compact for all $$x\in {\mathcal{X}}$$. Now, we show that Φx is convex for each $$x\in {\mathcal{X}}$$. Let $$g, g^{\prime }\in \varPhi x$$. Choose $$f,f^{\prime }\in S_{\mathcal{F},x}$$ such that $$g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds)$$ and $$g'(t)=a_{\zeta }f^{\prime }(t)+b_{\zeta } \int _{0}^{t}f^{\prime }(s)\,ds)$$ for almost all $$t\in I$$. Let $$0\leq \lambda \leq 1$$. Then we have

$$\lambda g(t)+(1-\lambda )g^{\prime }(t)=a_{\zeta } \bigl(\lambda f(t)+(1- \lambda )f^{\prime }(t) \bigr) +b_{\zeta } \int _{0}^{t} \bigl(\lambda f(s)+(1- \lambda )f^{\prime }(s) \bigr)\,ds.$$

Since $$S_{\mathcal{F},x}$$ is convex, $$\lambda g+(1-\lambda )g^{\prime }\in \varPhi x$$. This completes the proof. □

Note that the fixed point set of Φ is equal to the set of solutions for the inclusion problem (2). Now by using some different conditions, we show that the set of solutions for the fractional integro-differential inclusion problem could be infinite dimensional.

### Theorem 16

Suppose that $$\eta \in L^{1}(I,\mathbb{R}^{+})$$, $$\mathcal{F}: I \times \mathbb{R}^{m+n+3}\to \mathcal{P}_{cv,cp}(\mathbb{R})$$ is a multivalued map such that the function $$t\vdash \mathcal{F}(t,x_{1},x _{2},\ldots ,x_{m+n+3})$$ is measurable,

$$H\bigl(\mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3}), \mathcal{F}(t,y_{1},y _{2},\ldots ,y_{m+n+3})\bigr) \leq \eta (t)\sum_{i=1}^{m+n+3} \vert x_{i}-y_{i} \vert$$

and $$\|\mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\|=\sup \{|f| : f \in \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\}\leq \eta (t)$$ for almost all $$t\in I$$ and $$\in x_{1},x_{2},\ldots ,x_{m+n+3}, y_{1}, y _{2}, y_{m+n+3} \in \mathbb{R}$$. If Lebesgue measure of the set

$$\bigl\{ t: \dim \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})< 1 \textit{ for some } x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}\bigr\}$$

is zero and $$\Delta <1$$, then the set of all solutions for problem (2) is infinite dimensional, where $$\Delta =\eta ^{\ast } (1+n+\zeta _{0}+\iota _{0}+\sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} ) (1+n + \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )$$.

### Proof

Similar to Lemma 15, define the multivalued map $$\varPhi :{\mathcal{X}} \to \mathcal{P}({\mathcal{X}})$$ by

$$\varPhi (x)= \biggl\{ g\in {\mathcal{X}}: ~\text{there is} ~f\in S_{ \mathcal{F},x} ~\text{such that}~ g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds ~\text{for all}~ t\in I \biggr\} .$$

By using Lemma 15, $$\varPhi x\in \mathcal{P}_{cp,cv}({\mathcal{X}})$$ for all $$x\in \mathcal{X}$$. By using a similar proof in Theorem 12, we can prove that Φ is a contractive multivalued map. Now, we show that $$\dim \varPhi x>k$$ for all $$x\in \mathcal{X}$$ and $$k\geq 1$$. Let $$k\geq 1$$, $$x\in {\mathcal{X}}$$, and

\begin{aligned} \mathcal{G}(t)= {}& \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{1}}x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{2}}x(t), \dots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{m}}x(t), \\ & {} {}^{cF} \mathcal{I}^{\gamma _{1}}x(t), {}^{cF} \mathcal{I}^{ \gamma _{2}}x(t),\ldots ,{}^{cF}\mathcal{I}^{\gamma _{n}}x(t) \bigr) \end{aligned}

for all $$t\in I$$. By using Lemma 9, there are linearly independent measurable selections $$g_{1},\dots ,g_{k}$$ for $$\mathcal{G}$$. Consider the maps $$h_{i}(t)=a_{\zeta }g_{i}(t)+b_{ \zeta }(t)\int _{0}^{t}g_{i}(s)\,ds$$ for $$i=1,\dots ,k$$. Assume that $$\sum_{i=1}^{k} a_{i} h_{i}(t)=0$$ for almost $$t\in I$$. Since $$a_{\zeta },b_{\zeta }\neq 0$$, by using the Caputo–Fabrizio derivatives, we get $$\sum_{i=1}^{k} a_{i} g_{i}(t)=0$$ for almost $$t\in I$$. Hence, $$a_{1}=\cdots =a_{k}=0$$. This implies that $$h_{1}, \dots ,h_{k}$$ are linearly independent, and so $$\dim \varPhi x\geq k$$. Hence, we conclude that the set of fixed points of Φ is infinite dimensional by using Theorem 10. Thus, the set of all solutions for problem (2) is infinite dimensional. □

## 3 Conclusion

We guess that researchers will review different more fractional integro-differential inclusions in the near future. In this manuscript, we first investigate the existence of solutions for four fractional integro-differential inclusions including the new Caputo–Fabrizio derivation which has been introduced recently. Also, we show that dimension of the set of solutions for the second fractional integro-differential inclusion problem is infinite dimensional under some different conditions.