1 Introduction

It is well known that fractional integral, differential, and integro-differential equations have a great role in applications of various areas in physic, engineering, medicine, economy, and so on [24, 28, 42]. For example, in 2010, Tarasov [41] dealt with some applications of the fractional calculus to complex physical systems. Also Mainardi [32] presented comprehensively applications of the fractional calculus in dynamics of viscoelasticity. Then, Magin et al. [31] focused on the applications of the fractional calculus in signal processing. Furthermore, the reader can benefit from the papers; Gaul et al. [25], Samko et al. [39], Podlubny [37], Shaw et al. [40], Kilbas et al. [29], Das [21, 22], Hilfer [27], and Caponetto et al. [15] for the more informations about the fractional calculus. In addition to all these, nonlinear integral equations of fractional order have found wide applications in modeling of nonlinear problems and numerous research papers discussed integral equations of fractional order have been published [1, 23, 34, 43, 44].

In 1930, Kuratowski [30] presented the concept of measure of noncompactness. Furthermore, Gohberg et al. [26] defined the Hausdorff measure of noncompactness using \(\varepsilon \)-net. On the other hand, in 1955, Darbo [16] introduced a fixed-point theorem associated with the measure of noncompactness, and then, many different authors dealt with this fixed-point theorem and measure of noncompactness with its some applications [2, 33].

As is known, fixed-point theorems are used to show the existence of solutions of some equations. Therefore, fixed-point theorem has an important role in the theory of integral, differential, and integro-differential equations. In [2, 35, 38], the reader can find some fixed-point theorems.

Especially in recent years, solvability of integral, differential, and integro-differential equations of fractional order has been considered by many authors [3, 5,6,7, 9,10,11,12,13, 17,18,20, 36]. For example, the Cauchy problem for fractional differential equations:

$$\begin{aligned} \left\{ \begin{array}{l} ^{c}D^{\alpha }x(t)=f\left( t,x\left( t\right) \right) \\ \quad x\left( 0\right) =0\quad \quad \qquad \qquad \end{array} \right. \end{aligned}$$

has been handled by some authors in recent times, such that \(0<\alpha <1\) and x take values in any closed and bounded interval of \( {\mathbb {R}} \). Aghajani et al. [4] discussed the solvability of this equation under some sufficient conditions in the space of continuous functions which defined on interval \(\left[ 0,T\right] \) and take values in a Banach space E.

In this paper, we will consider the following integro-differential equation in the space of continuous functions which defined on interval \(\left[ 0,a \right] \) and take values in a Banach space E:

$$\begin{aligned} \left\{ \begin{array}{l} ^{c}D^{\alpha }x(t)=f\left( t,I^{\beta }x\left( t\right) \right) \\ x\left( 0\right) =0\quad \quad \qquad \qquad \end{array} \right. \end{aligned}$$
(1.1)

where \(t\in \left[ 0,a\right] \) and \(0<\alpha ,\beta \le 1.\)

It should be noted that for \(\beta =1\), integro-differential equation (1.1) transforms to the Cauchy problem for the fractional differential equation. Similarly, a fractional integral equation can be obtained from integro-differential equation (1.1) for \( \alpha =1\).

In Sect. 2, we present some definitions and preliminary results about the fractional calculus and the concept of Hausdorff measure of noncompactness. In the last section, we give our main result concerning the existence of solutions of the integro-differential equation  (1.1) by applying generalized Darbo fixed-point theorem associated with the measure of noncompactness given by [2], and we establish an example to show that our result is applicable.

2 Definitions and Auxiliary Facts

Suppose that \(\left( E,\left\| .\right\| \right) \) be an infinite-dimensional Banach space with zero element \(\theta .\) We write \(B\left( x,r\right) \) and \(B\left[ x,r\right] \) to denote the open and closed ball centered at x with radius r,  respectively. Especially, we write \(B_{r} \) instead of \(B\left[ \theta ,r\right] \). We write \({\overline{X}}\), \( Conv\, X\) to denote the closure of X and convex closure of X , respectively. Moreover, let \({\mathfrak {M}}_{E}\) indicates the family of all nonempty bounded subsets of E and \({\mathfrak {N}}_{E}\) indicates its subfamily of all relatively compact sets. Finally, the standard algebraic operations on sets are denoted by \(\lambda X\) and \(X+Y\), respectively.

Denote by \(C\left( J,E\right) \) the space of all continuous functions from \( J=\left[ a,b\right] \) to Banach space E with the norm:

$$\begin{aligned} \left\| x\right\| _{C\left( J,E\right) }=\sup _{s\in J}\left\| x\left( s\right) \right\| _{E}. \end{aligned}$$

Also suppose that \(L^{1}\left( J,E\right) \) is the space of Bochner integrable functions with the norm:

$$\begin{aligned} \left\| x\right\| _{L^{1}\left( J,E\right) }=\int _{a}^{b}\left\| x\left( s\right) \right\| _{E}{\text {d}}s. \end{aligned}$$

We would like to indicate that if there is not any confusion, we write only \( \left\| x\right\| \) instead of \(\left\| x\right\| _{C\left( J,E\right) }\), \(\left\| x\right\| _{L^{1}\left( J,E\right) }\) or \( \left\| x\right\| _{E}.\)

Let us remember the concept of fractional integrals and derivatives.

Definition 2.1

[27, 29] Let x be a function defined on J and \(\alpha >0\); then

$$\begin{aligned} I_{a^{+}}^{\alpha }x(t)=\frac{1}{\Gamma (\alpha )}\int _{a}^{t}\frac{x(s)}{ (t-s)^{1-\alpha }}{\text {d}}s \end{aligned}$$

is called the Riemann–Liouville fractional integral of order \(\alpha \) of function x, where symbol of \(\Gamma \) denote the gamma function defined by:

$$\begin{aligned} \Gamma (\alpha )=\int _{0}^{\infty }t^{\alpha -1}e^{-t}{\text {d}}t. \end{aligned}$$

Definition 2.2

[29] Let x be a function defined on J and \(0<\alpha <1\); then

$$\begin{aligned} D_{a^{+}}^{\alpha }x(t)=\frac{1}{\Gamma (1-\alpha )}\frac{{\text {d}}}{{\text {d}}t}\left( \int _{a}^{t}\frac{x(s)}{(t-s)^{\alpha }}{\text {d}}s\right) \end{aligned}$$

is called the Riemann–Liouville fractional derivative of order \(\alpha \) of the function x.

Definition 2.3

[29] For a function x defined on J, the Caputo fractional derivative of order \(\alpha \in \left( 0,1\right) \) is defined by:

$$\begin{aligned} ^{c}D_{a^{+}}^{\alpha }x(t)=D_{a^{+}}^{\alpha }\left( x(t)-x\left( a\right) \right) . \end{aligned}$$

We write \(I^{\alpha }x(t)\) and \(^{c}D^{\alpha }x(t)\) instead of the notations \(I_{0^{+}}^{\alpha }x(t)\) and \(^{c}D_{0^{+}}^{\alpha }x(t),\) respectively.

Lemma 2.4

[29, 39] Let x,  \(y\in L^{1}\left( J, {\mathbb {R}} \right) \) and \(\alpha ,\)\(\beta >0.\) Then, we have:

  1. (1)

    \(I_{a^{+}}^{\alpha }I_{a^{+}}^{\beta }x(t)=I_{a^{+}}^{\beta }I_{a^{+}}^{\alpha }x(t)=I_{a^{+}}^{\alpha +\beta }x(t),\)

  2. (2)

    \(I_{a^{+}}^{\alpha }\left( x(t)+y\left( t\right) \right) =I_{a^{+}}^{\alpha }x(t)+I_{a^{+}}^{\alpha }y(t),\)

  3. (3)

    \(I_{a^{+}}^{\alpha }\)\(^{c}D_{a^{+}}^{\alpha }x(t)=x\left( t\right) -x\left( a\right) ,\)\(\alpha \in \left( 0,1\right) ,\)

  4. (4)

    \(^{c}D_{a^{+}}^{\alpha }I_{a^{+}}^{\alpha }x(t)=x\left( t\right) ,\)\( \alpha \in \left( 0,1\right) ,\)

  5. (5)

    \(^{c}D_{a^{+}}^{\alpha }x(t)=I_{a^{+}}^{1-\alpha }x^{\prime }\left( t\right) ,\)\(\alpha \in \left( 0,1\right) ,\)

  6. (6)

    \(^{c}D_{a^{+}}^{\alpha }\)\(^{c}D_{a^{+}}^{\beta }x(t)\ne \)\( ^{c}D_{a^{+}}^{\beta }\)\(^{c}D_{a^{+}}^{\alpha }x(t),\)\(\alpha ,\)\(\beta \in \left( 0,1\right) ,\)

  7. (7)

    \(^{c}D_{a^{+}}^{\alpha }\)\(^{c}D_{a^{+}}^{\beta }x(t)\ne \)\( ^{c}D_{a^{+}}^{\alpha +\beta }x(t),\)\(\alpha ,\)\(\beta ,\)\(\alpha +\beta \in \left( 0,1\right) .\)

We will use the Hausdorff measure of noncompactness \(\gamma \) defined as follows.

Definition 2.5

[26] Let \(\left( E,d\right) \) be a metric space. Then, a mapping \( \gamma :{\mathfrak {M}}_{E}\rightarrow {\mathbb {R}} _{+}=\left[ 0,\infty \right) \) defined by:

$$\begin{aligned} \gamma \left( A\right) =\inf \left\{ \varepsilon >0:A\subset \bigcup _{k=1}^{n}B\left( x_{k},r_{k}\right) ,\text { }x_{k}\in E\text { and } r_{k}<\varepsilon \right\} \end{aligned}$$

is said to be Hausdorff measure of noncompactness in E.

Actually, this definition means that \(\gamma \left( A\right) \) is the infimum of all \(\varepsilon \) for which A has an \(\varepsilon \)-finite net.

Lemma 2.6

[8] Let E be a real Banach space and A,  \(B\in \mathfrak {M }_{E}.\) Then, the following properties are satisfied:

  1. (1)

    \(\gamma \left( A\right) =0\) iff \({\overline{A}}\) is compact;

  2. (2)

    \(\gamma \left( A\right) =\gamma \left( {\overline{A}}\right) =\gamma \left( Conv\, A\right) ;\)

  3. (3)

    \(A\subseteq B\) implies \(\gamma \left( A\right) \le \gamma \left( B\right) ;\)

  4. (4)

    \(\gamma \left( \alpha A\right) =\left| \alpha \right| \gamma \left( A\right) ,\) for all \(\alpha \in {\mathbb {R}};\)

  5. (5)

    \(\gamma \left( A\cup B\right) =\max \left\{ \gamma \left( A\right) ,\gamma \left( B\right) \right\} ;\)

  6. (6)

    \(\gamma \left( A+B\right) \le \gamma \left( A\right) +\gamma \left( B\right) ;\)

  7. (7)

    If \(\left( A_{n}\right) \) is a decreasing sequence of closed, nonempty sets in \({\mathfrak {M}}_{E}\) and \(\lim _{n\rightarrow \infty }\gamma \left( A_{n}\right) =0,\) then \(\cap _{n=1}^{\infty }A_{n}\) is nonempty and compact in E.

Definition 2.7

[4] A mapping \(F:C\subseteq E\rightarrow E\) is said to be \( \gamma \)-contraction on C if there exists a constant \(k\in \left( 0,1\right) \), such that the inequality

$$\begin{aligned} \gamma \left( FX\right) \le k\gamma \left( X\right) \end{aligned}$$

is satisfied for all bounded and closed subset X of C.

Theorem 2.8

[2] Let C be a nonempty, closed, bounded, and convex subset of a Banach space E and \(F:C\rightarrow C\) be a continuous mapping. Assume that there exists a monoton and increasing function \(\phi :\left[ 0,\infty \right) \)\(\rightarrow \left[ 0,\infty \right) \), such that \(\lim _{n\rightarrow \infty }\phi ^{n}\left( t\right) =0\) for all \(t\in \left[ 0,\infty \right) \) and the inequality:

$$\begin{aligned} \gamma \left( FX\right) \le \phi \left( \gamma \left( X\right) \right) \end{aligned}$$
(2.1)

holds for each \(X\subseteq C\), where \(\gamma \) is Hausdorff measure of noncompactness in E. Then, F has at least one fixed point in C.

Lemma 2.9

[8] If \(X\subseteq C\left( J,E\right) \) is a bounded and equicontinuous set, then \(\gamma \left( X\left( t\right) \right) \) is continuous on J and

$$\begin{aligned} \gamma \left( X\right) =\sup _{t\in J}\gamma \left( X\left( t\right) \right) ,\qquad \gamma \left( \left\{ \int _{0}^{t}x\left( s\right) {\text {d}}s:x\in X\right\} \right) \le \int _{0}^{t}\gamma \left( X\left( s\right) \right) {\text {d}}s. \end{aligned}$$

Lemma 2.10

[14] Let X be a bounded set, and then, for each \( \varepsilon >0\), there exists a sequence \(\left( x_{n}\right) \subseteq X\), such that

$$\begin{aligned} \gamma \left( X\right) \le 2\gamma \left( \left\{ x_{n}:n\ge 1\right\} \right) +\varepsilon . \end{aligned}$$

Lemma 2.11

[4] If \(\left\{ x_{n}:n\ge 1\right\} \subseteq L^{1}\left( J,E\right) \) satisfies \(\left\| x_{n}\left( t\right) \right\| \le p\left( t\right) \) a.e. on J for all \(n\ge 1\) with some \(p\in \)\(L^{1}\left( J, {\mathbb {R}} _{+}\right) ,\) then \(\gamma \left( \left\{ x_{n}\left( t\right) :n\ge 1\right\} \right) \)\(\in \)\(L^{1}\left( J, {\mathbb {R}} _{+}\right) \) and the following inequality holds:

$$\begin{aligned} \gamma \left( \left\{ \int _{0}^{t}x_{n}\left( s\right) {\text {d}}s:n\ge 1\right\} \right) \le 2\int _{0}^{t}\gamma \left( \left\{ x_{n}\left( s\right) :n\ge 1\right\} \right) {\text {d}}s. \end{aligned}$$

Lemma 2.12

A function \(x\in C\left( J,E\right) \) is a solution of Eq. (1.1) if and only if x is a solution of the fractional integral equation:

$$\begin{aligned} x\left( t\right) =\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s, \end{aligned}$$
(2.2)

where J is considered as \(\left[ 0,a\right] .\)

Proof

Let x be a solution of Eq.  (1.1) that is \( ^{c}D^{\alpha }x(t)=f\left( t,I^{\beta }x\left( t\right) \right) \) and \( x\left( 0\right) =0\). Then, using Lemma 2.4 and taking into account that \(f\left( \cdot ,x\right) \in C\left( J,E\right) \), we get:

$$\begin{aligned} I^{\alpha \text { }c}D^{\alpha }x(t)=I^{\alpha }f\left( t,I^{\beta }x\left( t\right) \right) \end{aligned}$$

and so

$$\begin{aligned} x\left( t\right) -x\left( 0\right) =\frac{1}{\Gamma (\alpha )} \int _{0}^{t}\left( t-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s. \end{aligned}$$

Since \(x\left( 0\right) =0,\)x satisfies Eq. (2.2).

On the other hand, if x is a solution of Eq. (2.2), then we can write:

$$\begin{aligned} ^{c}D^{\alpha }x(t)= & {} \text { }^{c}D^{\alpha }\left( \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right) \\= & {} \text { }^{c}D^{\alpha }\left( I^{\alpha }f\left( \cdot ,I^{\beta }x\right) \right) \left( t\right) \\= & {} \text { }^{c}D^{\alpha }I^{\alpha }f\left( t,I^{\beta }x\left( t\right) \right) \\= & {} f\left( t,I^{\beta }x\left( t\right) \right) \end{aligned}$$

from Lemma 2.4. Also it easy to see that \(x\left( 0\right) =0\) from Eq. (2.2). Therefore, x is a solution of Eq. (1.1) . \(\square \)

It should be noted that since the function x is an abstract function, the integrals which appear above are considered Bochner’s sense.

3 The Main Result

We study the solvability of integro-differential equation (1.1) with the following conditions, such that \(J=\left[ 0,a\right] .\)

\({\mathbf {(a}}_{1}{\mathbf {)}}\):

The function \(f:J\times E\rightarrow E\) satisfies Carathéodory conditions, that is, \(f\left( t,x\right) \) is a measurable function of t for each fixed x and \(f\left( t,x\right) \) is a continuous function of x for a.e. \(t\in J.\)

\({\mathbf {(a}}_{2}{\mathbf {)}}\):

There exists a non-decreasing and continuous function h :  \( {\mathbb {R}} _{+}\rightarrow {\mathbb {R}} _{+}\), such that the inequality

$$\begin{aligned} \left\| f\left( t,x\right) \right\| \le h(\left\| x\right\| ) \end{aligned}$$

holds for all \(x\in C\left( J,E\right) \) and \(t\in J.\)

\({\mathbf {(a}}_{3}{\mathbf {)}}\):

The inequality

$$\begin{aligned} \frac{a^{\alpha }h\left( \frac{a^{\beta }r}{\Gamma (\beta +1)}\right) }{ \Gamma (\alpha +1)}\le r \end{aligned}$$
(3.1)

has a positive solution \(r_{0}\).

\({\mathbf {(a}}_{4}{\mathbf {)}}\):

\(\phi :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is a non-decreasing map, such that \( \lim _{n\rightarrow \infty }\phi ^{n}\left( t\right) =0,\) for all \(t\in {\mathbb {R}} _{+}.\) Also, there exist a function \(L\in L^{1}\left( J, {\mathbb {R}} _{+}\right) \), such that the inequality

$$\begin{aligned} \gamma \left( f\left( t,B\right) \right) \le L\left( t\right) \phi \left( \gamma \left( B\right) \right) \end{aligned}$$
(3.2)

holds for all nonempty subset B of \(B_{r_{0}}\) and

$$\begin{aligned} \max \left\{ \frac{a^{\beta }}{\Gamma (\beta +1)},\frac{4\left\| L\right\| a^{\alpha }}{\Gamma (\alpha +1)}\right\} <1. \end{aligned}$$
(3.3)

Theorem 3.1

Under assumptions \({\mathbf {(a}}_{1}\mathbf {)-(a}_{4}{\mathbf {)}}\), Eq. (1.1) has at least one solution \(x=x(t)\) which belongs to \(B_{r_{0}}\subset C\left( J,E\right) .\)

Proof

Let we define operator F as:

$$\begin{aligned} \left( Fx\right) \left( t\right) =\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-s)^{\alpha -1}f(s,I^{\beta }x\left( s\right) ){\text {d}}s. \end{aligned}$$

It is easy to see that F is well defined and every fixed point of F is a solution of Eq. (2.2) .

Now, we show that operator F has at least one fixed point on \(C\left( J,E\right) \) under assumption \({\mathbf {(a}}_{1}\mathbf {)-(a}_{4}\mathbf {).}\)

Let \(\varepsilon >0.\) If we choose

$$\begin{aligned} \delta =\left( \frac{\varepsilon \Gamma (\alpha +1)}{2h\left( \frac{a^{\beta }r_{0}}{\Gamma (\beta +1)}\right) }\right) ^{\frac{1}{\alpha }}, \end{aligned}$$

then for every \(x\in B_{r_{0}}\subset C\left( J,E\right) \) and \( t_{1},t_{2}\in J\), such that \(\left| t_{1}-t_{2}\right| \le \delta \) (without loss of generality, we assume \(t_{2}\le t_{1}\)), we can write:

$$\begin{aligned}&\left\| \left( Fx\right) \left( t_{1}\right) -\left( Fx\right) \left( t_{2}\right) \right\| \nonumber \\&\quad =\frac{1}{\Gamma (\alpha )}\left\| \int _{0}^{t_{1}}\left( t_{1}-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right. \nonumber \\&\qquad -\left. \int _{0}^{t_{2}}\left( t_{2}-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right\| \nonumber \\&\quad \le \frac{1}{\Gamma (\alpha )}\left\{ \left\| \int _{0}^{t_{2}}\left[ \left( t_{1}-s\right) ^{\alpha -1}-\left( t_{2}-s\right) ^{\alpha -1}\right] f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right\| \right. \nonumber \\&\qquad \left. +\left\| \int _{t_{2}}^{t_{1}}\left( t_{1}-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right\| \right\} \nonumber \\&\quad \le \frac{1}{\Gamma (\alpha )}\left\{ \int _{0}^{t_{2}}\left[ \left( t_{2}-s\right) ^{\alpha -1}-\left( t_{1}-s\right) ^{\alpha -1}\right] \left\| f\left( s,I^{\beta }x\left( s\right) \right) \right\| {\text {d}}s\right. \nonumber \\&\qquad \left. +\int _{t_{2}}^{t_{1}}\left( t_{1}-s\right) ^{\alpha -1}\left\| f\left( s,I^{\beta }x\left( s\right) \right) \right\| {\text {d}}s\right\} \nonumber \\&\quad \le \frac{h\left( \left\| I^{\beta }x\left( s\right) \right\| \right) }{\Gamma (\alpha )}\left\{ \int _{0}^{t_{2}}\left[ \left( t_{2}-s\right) ^{\alpha -1}-\left( t_{1}-s\right) ^{\alpha -1}\right] {\text {d}}s\right. \nonumber \\&\qquad +\left. \int _{t_{2}}^{t_{1}}\left( t_{1}-s\right) ^{\alpha -1}{\text {d}}s\right\} . \end{aligned}$$
(3.4)

Also we have

$$\begin{aligned} \left\| I^{\beta }x\left( s\right) \right\|= & {} \left\| \frac{1}{ \Gamma (\beta )}\int _{0}^{s}(s-r)^{\beta -1}x\left( r\right) {\text {d}}r\right\| \nonumber \\\le & {} \frac{\left\| x\right\| }{\Gamma (\beta )}\int _{0}^{s}(s-r)^{\beta -1}{\text {d}}r \nonumber \\\le & {} \frac{a^{\beta }\left\| x\right\| }{\Gamma (\beta +1)}. \end{aligned}$$
(3.5)

If we consider (3.4) with (3.5), we conclude:

$$\begin{aligned} \left\| \left( Fx\right) \left( t_{1}\right) -\left( Fx\right) \left( t_{2}\right) \right\|\le & {} \frac{h\left( \frac{a^{\beta }\left\| x\right\| }{\Gamma (\beta +1)}\right) }{\Gamma (\alpha +1)}\left[ 2\left( t_{1}-t_{2}\right) ^{\alpha }+t_{2}^{\alpha }-t_{1}^{\alpha }\right] \\\le & {} \frac{h\left( \frac{a^{\beta }r_{0}}{\Gamma (\beta +1)}\right) }{ \Gamma (\alpha +1)}2\left( t_{1}-t_{2}\right) ^{\alpha } \\\le & {} \frac{2h\left( \frac{a^{\beta }r_{0}}{\Gamma (\beta +1)}\right) \delta ^{\alpha }}{\Gamma (\alpha +1)} \\= & {} \varepsilon . \end{aligned}$$

Therefore, F is an operator from \(B_{r_{0}}\) into \(C\left( J,E\right) \) and \( FB_{r_{0}}\) is equicontinuous. Moreover, by taking this situation into account, we obtain:

$$\begin{aligned} \gamma \left( X\right) =\sup _{t\in J}\gamma \left( X\left( t\right) \right) \end{aligned}$$

and

$$\begin{aligned} \gamma \left( \int _{0}^{t}X\left( s\right) {\text {d}}s\right) \le \int _{0}^{t}\gamma \left( X\left( s\right) \right) {\text {d}}s. \end{aligned}$$

for any nonempty subset X of \(FB_{r_{0}}.\)

On the other hand for any \(x\in B_{r_{0}},\) we have:

$$\begin{aligned} \left\| \left( Fx\right) \left( t\right) \right\|= & {} \left\| \frac{1 }{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}f(s,I^{\beta }x\left( s\right) ){\text {d}}s\right\| \nonumber \\\le & {} \frac{1}{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}\left\| f(s,I^{\beta }x\left( s\right) )\right\| {\text {d}}s \nonumber \\\le & {} \frac{1}{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}h\left( \left\| I^{\beta }x\left( s\right) \right\| \right) {\text {d}}s. \end{aligned}$$
(3.6)

If we consider (3.5) with (3.6) in addition to (3.1), we obtain:

$$\begin{aligned} \left\| \left( Fx\right) \left( t\right) \right\|\le & {} \frac{1}{ \Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}h\left( \frac{a^{\beta }\left\| x\right\| }{\Gamma (\beta +1)}\right) {\text {d}}s \\\le & {} h\left( \frac{a^{\beta }\left\| x\right\| }{\Gamma (\beta +1)} \right) \frac{a^{\alpha }}{\Gamma (\alpha +1)} \\\le & {} h\left( \frac{a^{\beta }r_{0}}{\Gamma (\beta +1)}\right) \frac{ a^{\alpha }}{\Gamma (\alpha +1)} \\\le & {} r_{0}. \end{aligned}$$

Therefore, \(F:B_{r_{0}}\rightarrow B_{r_{0}}.\)

Now, we prove continuity of the operator F. To do this, let \(\left( x_{n}\right) \subset B_{r_{0}}\) be a sequence which converges to \(x\in B_{r_{0}}.\) Then, we have:

$$\begin{aligned}&\left\| \left( Fx_{n}\right) \left( t\right) -\left( Fx\right) \left( t\right) \right\| \\&\quad =\left\| \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}f\left( s,I^{\beta }x_{n}\left( s\right) \right) {\text {d}}s-\frac{1}{ \Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}f\left( s,I^{\beta }x\left( s\right) \right) {\text {d}}s\right\| \\&\quad =\frac{1}{\Gamma (\alpha )}\left\| \int _{0}^{t}\left( t-s\right) ^{\alpha -1}\left[ f\left( s,I^{\beta }x_{n}\left( s\right) \right) -f\left( s,I^{\beta }x\left( s\right) \right) \right] {\text {d}}s\right\| . \end{aligned}$$

Furthermore, sake for brevity, let us define a sequence \(\left( y_{n}\right) \) as:

$$\begin{aligned} y_{n}\left( t\right) =\left( I^{\beta }x_{n}\right) \left( t\right) =\frac{1 }{\Gamma (\beta )}\int _{0}^{t}(t-s)^{\beta -1}x_{n}\left( s\right) , \end{aligned}$$

for \(\left( x_{n}\right) \subseteq B_{r_{0}}.\) It is easy to see that operator \(I^{\beta }\) is continuous on \(B_{r_{0}}\) and so \(\left( y_{n}\right) \rightarrow y\), such that \(y=I^{\beta }x.\) We would like to indicate immediately that, for any \(\varepsilon >0\) and for any fixed \(n\ge 1,\) we get:

$$\begin{aligned} \left\| y_{n}\left( t_{1}\right) -y_{n}\left( t_{2}\right) \right\|= & {} \frac{1}{\Gamma (\beta )}\left\| \int _{0}^{t_{1}}(t_{1}-s)^{\beta -1}x_{n}\left( s\right) {\text {d}}s-\int _{0}^{t_{2}}(t_{2}-s)^{\beta -1}x_{n}\left( s\right) {\text {d}}s\right\| \\\le & {} \frac{\left\| x_{n}\right\| }{\Gamma (\beta )}\left( \int _{0}^{t_{1}}(t_{1}-s)^{\beta -1}{\text {d}}s-\int _{0}^{t_{2}}(t_{2}-s)^{\beta -1}{\text {d}}s\right) \\\le & {} \frac{\left\| x_{n}\right\| \left( t_{1}^{\beta }-t_{2}^{\beta }\right) }{\Gamma (\beta +1)} \\\le & {} \frac{\left\| x_{n}\right\| \left( t_{1}-t_{2}\right) ^{\beta } }{\Gamma (\beta +1)} \\\le & {} \frac{\delta \left\| x_{n}\right\| }{\Gamma (\beta +1)} \\\le & {} \frac{\delta r_{0}}{\Gamma (\beta +1)} \\= & {} \varepsilon , \end{aligned}$$

where \(t_{1},t_{2}\in J\) with \(\left| t_{1}-t_{2}\right| \le \delta \), such that

$$\begin{aligned} \delta =\frac{\varepsilon \Gamma (\beta +1)}{r_{0}}. \end{aligned}$$

Therefore, \(\left( y_{n}\right) \subseteq C\left( J,E\right) \) and \(\left\{ y_{n}:n\ge 1\right\} \) is equicontinuous. Also by taking into account (3.3) and (3.5), we obtain \(\left( y_{n}\right) \subseteq B_{r_{0}}.\)

Thus, we have:

$$\begin{aligned}&\left\| \left( Fx_{n}\right) \left( t\right) -\left( Fx\right) \left( t\right) \right\| \\&\quad \le \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}\left\| f\left( s,y_{n}\left( s\right) \right) -f\left( s,y\left( s\right) \right) \right\| {\text {d}}s. \end{aligned}$$

On the other hand, we conclude:

$$\begin{aligned} \left\| f\left( s,y_{n}\left( s\right) \right) -f\left( s,y\left( s\right) \right) \right\| \rightarrow 0 \end{aligned}$$

as \(n\rightarrow \infty ,\) from Carathéodory continuity of the function f. Also, we can write:

$$\begin{aligned} \left\| f\left( s,y_{n}\left( s\right) \right) -f\left( s,y\left( s\right) \right) \right\|\le & {} \left\| f\left( s,y_{n}\left( s\right) \right) \right\| +\left\| f\left( s,y\left( s\right) \right) \right\| \\\le & {} h\left( \left\| y_{n}\left( s\right) \right\| \right) +h\left( \left\| y\left( s\right) \right\| \right) \end{aligned}$$

from \(({\mathbf {a}}_{2})\) and if we define the function g as

$$\begin{aligned} g\left( s\right) =\left( t-s\right) ^{\alpha -1}\left[ h\left( \left\| y_{n}\left( s\right) \right\| \right) +h\left( \left\| y\left( s\right) \right\| \right) \right] , \end{aligned}$$

g is a Lebesgue integrable function on J. Then, we have:

$$\begin{aligned} \left\| \left( Fx_{n}\right) \left( t\right) -\left( Fx\right) \left( t\right) \right\| \le \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\left( t-s\right) ^{\alpha -1}\left\| f\left( s,y_{n}\left( s\right) \right) -f\left( s,y\left( s\right) \right) \right\| {\text {d}}s\rightarrow 0 \end{aligned}$$

as \(n\rightarrow \infty .\) This means that operator \(F:B_{r_{0}}\rightarrow B_{r_{0}}\) is continuous.

To use Theorem 2.8, we have to show that operator F satisfies the inequality (2.1) .

From Lemma 2.10, there exists a sequence \(\left( x_{n}\right) \subset B_{r_{0}}\), such that the following inequality holds:

$$\begin{aligned} \gamma \left( \left( FB_{r_{0}}\right) \left( t\right) \right)= & {} \gamma \left( \left\{ \frac{1}{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}f(s,I^{\beta }x\left( s\right) ){\text {d}}s:x\in B_{r_{0}}\right\} \right) \\\le & {} \frac{2}{\Gamma (\alpha )}\gamma \left( \left\{ \int _{0}^{t}(t-s)^{\alpha -1}f(s,I^{\beta }x_{n}\left( s\right) ){\text {d}}s:n\ge 1\right\} \right) +\varepsilon . \end{aligned}$$

Also using Lemma 2.11 and the condition \(\left( {\mathbf {a}} _{4}\right) \), we get:

$$\begin{aligned} \gamma \left( \left( FB_{r_{0}}\right) \left( t\right) \right)\le & {} \frac{4 }{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}\gamma \left( \left\{ f(s,I^{\beta }x_{n}\left( s\right) ):n\ge 1\right\} \right) {\text {d}}s+\varepsilon \nonumber \\= & {} \frac{4}{\Gamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}\gamma \left( \left\{ f(s,y_{n}\left( s\right) ):n\ge 1\right\} \right) {\text {d}}s+\varepsilon \nonumber \\\le & {} \frac{4\phi \left( \gamma \left( \left\{ y_{n}\left( s\right) :n\ge 1,\text { }0\le s\le t\right\} \right) \right) }{\Gamma (\alpha )} \int _{0}^{t}(t-s)^{\alpha -1}L\left( s\right) {\text {d}}s+\varepsilon \nonumber \\\le & {} \frac{4\left\| L\right\| a^{\alpha }\phi \left( \gamma \left( \left\{ y_{n}\left( s\right) :n\ge 1,\text { }0\le s\le t\right\} \right) \right) }{\Gamma (\alpha +1)}+\varepsilon . \end{aligned}$$
(3.7)

On the other hand, from equicontinuity of \(\left\{ y_{n}:n\ge 1\right\} ,\) we derive:

$$\begin{aligned} \gamma \left( \left\{ y_{n}:n\ge 1\right\} \right) =\sup _{t\in J}\gamma \left( \left\{ y_{n}\left( t\right) :n\ge 1\right\} \right) =\sup _{t\in J}\left( \sup _{s\in \left[ 0,t\right] }\gamma \left( \left\{ y_{n}\left( s\right) :n\ge 1\right\} \right) \right) . \end{aligned}$$

In the light of this fact, taking supremum over \(t\in J\) in (3.7), we obtain:

$$\begin{aligned} \gamma \left( FB_{r_{0}}\right) \le \frac{4\left\| L\right\| a^{\alpha }}{\Gamma (\alpha +1)}\phi \left( \gamma \left( \left\{ y_{n}:n\ge 1\right\} \right) \right) +\varepsilon . \end{aligned}$$
(3.8)

Moreover, we can write:

$$\begin{aligned} \left\| y_{n}\left( t\right) \right\| =\left\| \frac{1}{\Gamma (\beta )}\int _{0}^{t}(t-s)^{\beta -1}x_{n}\left( s\right) {\text {d}}s\right\| \le \frac{r_{0}a^{\beta }}{\Gamma (\beta +1)} \end{aligned}$$

for \(\left( x_{n}\right) \subseteq B_{r_{0}}\) and using (3.3), we have \(\left\{ y_{n}:n\ge 1\right\} \subseteq B_{r_{0}}\) and:

$$\begin{aligned} \gamma \left( \left\{ y_{n}:n\ge 1\right\} \right) \le \gamma \left( B_{r_{0}}\right) . \end{aligned}$$

If we consider this result together with (3.8), then we get:

$$\begin{aligned} \gamma \left( FB_{r_{0}}\right)\le & {} \frac{4\left\| L\right\| a^{\alpha }}{\Gamma (\alpha +1)}\phi \left( \gamma \left( \left\{ y_{n}:n\ge 1\right\} \right) \right) +\varepsilon \nonumber \\\le & {} \frac{4\left\| L\right\| a^{\alpha }}{\Gamma (\alpha +1)}\phi \left( \gamma \left( B_{r_{0}}\right) \right) +\varepsilon . \end{aligned}$$

Considering \(\varepsilon \rightarrow 0\), we conclude:

$$\begin{aligned} \gamma \left( FB_{r_{0}}\right) \le \frac{4\left\| L\right\| a^{\alpha }}{\Gamma (\alpha +1)}\phi \left( \gamma \left( B_{r_{0}}\right) \right) . \end{aligned}$$

Finally, taking the condition \(\left( {\mathbf {a}}_{4}\right) \) into account, we have:

$$\begin{aligned} \gamma \left( FB_{r_{0}}\right) \le \phi \left( \gamma \left( B_{r_{0}}\right) \right) . \end{aligned}$$

Thus, from Theorem 2.8 and Lemma 2.12, we can easily say that operator F has at least one fixed point on \(B_{r_{0}}\subset C\left( J,E\right) \) and this completes the proof. \(\square \)

Example 3.2

Consider the following integro-differential equation in \(C\left( J,c_{0}\right) \):

$$\begin{aligned} \left\{ \begin{array}{l} ^{c}D^{\frac{1}{2}}x(t)=f\left( t,I^{\frac{1}{3}}x\left( t\right) \right) \\ x\left( 0\right) =0\quad \quad \qquad \qquad \end{array} \right. , \end{aligned}$$
(3.9)

such that \(J=\left[ 0,0.049\right] \). Where, for every \(x=\left( x_{n}\right) _{n=1}^{\infty }\in c_{0}\) and \(t\in J,\) function \(f\left( t,x\right) \) is defined as:

$$\begin{aligned} f\left( t,x\right) =\left( \frac{n^{2}\ln \left( \left| y_{k\left( t\right) }^{n}\right| +1\right) +t}{2n^{2}\left( t^{2}+1\right) }\right) _{n=1}^{\infty }, \end{aligned}$$

such that \(y_{k\left( t\right) }=\left( y_{k\left( t\right) }^{n}\right) _{n=1}^{\infty }\in c_{0}\) and

$$\begin{aligned} y_{k\left( t\right) }^{n}=\left\{ \begin{array}{ll} 0, &{}\quad k\left( t\right) =n \\ x_{n}, &{}\quad k\left( t\right) \ne n \end{array} \right. . \end{aligned}$$

Also, we consider function \(k:J\rightarrow {\mathbb {N}} ^{+}\) as:

$$\begin{aligned} k\left( t\right) =\left\{ \begin{array}{ll} 1, &{}\quad t=0 \\ \left\lfloor \frac{1}{t}\right\rfloor , &{}\quad \text {otherwise} \end{array} \right. , \end{aligned}$$

where \(\left\lfloor t\right\rfloor \) denotes integer value of t.

From Lemma 2.12, it is well known that Eq. (3.9) has a solution if and only if the following equation has a solution in \( C\left( J,c_{0}\right) \):

$$\begin{aligned} x\left( t\right) =\frac{1}{\Gamma \left( \frac{1}{2}\right) } \int _{0}^{t}\left( t-s\right) ^{-\frac{1}{2}}f\left( s,\frac{1}{\Gamma \left( \frac{1}{3}\right) }\int _{0}^{s}\left( s-m\right) ^{-\frac{2}{3} }x\left( m\right) {\text {d}}m\right) {\text {d}}s. \end{aligned}$$
(3.10)

For any \(x\in c_{0}\) and \(t\in J\), we have:

$$\begin{aligned} \left\| f\left( t,x\right) \right\|= & {} \sup _{n\ge 1}\left| \frac{ n^{2}\ln \left( \left| y_{k\left( t\right) }^{n}\right| +1\right) +t }{2n^{2}\left( t^{2}+1\right) }\right| \\= & {} \frac{1}{2\left( t^{2}+1\right) }\sup _{n\ge 1}\left| \frac{n^{2}\ln \left( \left| y_{k\left( t\right) }^{n}\right| +1\right) +t}{n^{2}} \right| \\\le & {} \frac{1}{2\left( t^{2}+1\right) }\left( \sup _{n\ge 1}\left| y_{k\left( t\right) }^{n}\right| +t\right) \\\le & {} \frac{1}{2}\left( \left\| y_{k\left( t\right) }\right\| +t\right) \\\le & {} \frac{1}{2}\left( \left\| x\right\| +\frac{\pi }{64}\right) . \end{aligned}$$

Therefore, we can choose:

$$\begin{aligned} h\left( x\right) =\frac{1}{2}\left( x+\frac{\pi }{64}\right) . \end{aligned}$$

To verify assumption \({\mathbf {(a}}_{3}{\mathbf {)}}\), we observe that the inequality appearing in this assumption has the form:

$$\begin{aligned} \left( \frac{\root 3 \of {a}r}{2\Gamma \left( \frac{4}{3}\right) }+\frac{\pi }{ 128}\right) \frac{\sqrt{a}}{\Gamma \left( \frac{3}{2}\right) }\le , r \end{aligned}$$

where \(a=0.049.\) It is easy to verify that any number \(r_{0}\ge 0.00640965\) satisfies the above inequality.

Also, it can be seen that f satisfies the Carathéodory conditions in a similar way from [4].

To justify the inequality (3.2), we benefit from the advantage of the following theorem [33]:

Theorem 3.3

Let B be a bounded subset of the normed space X, where X is \(l_{p}\)\( \left( 1\le p<\infty \right) \) or \(c_{0}\). Also, \(P_{n}:X\rightarrow X\) be an operator defined by:

$$\begin{aligned} P_{n}\left( x_{1},x_{2},\ldots \right) =\left( x_{1},x_{2},\ldots ,x_{n},0,0,\ldots \right) \end{aligned}$$

for \(\left( x_{1},x_{2},\ldots \right) \in X\). Then, \(\gamma \left( B\right) =\lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) x\right\| .\)

Let B is an arbitrary nonempty subset of \(B_{r_{0}}\subseteq c_{0}\) and \( t\in J.\) Then, we get:

$$\begin{aligned} \gamma \left( f\left( t,B\right) \right)= & {} \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) f\left( t,x\right) \right\| _{\infty } \\= & {} \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( \frac{n^{2}\ln \left( \left| y_{k\left( t\right) }^{n}\right| +1\right) +t}{2n^{2}\left( t^{2}+1\right) }\right) _{n=1}^{\infty }\right\| \\\le & {} \frac{1}{2\left( t^{2}+1\right) }\left( \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( \ln \left( \left| y_{k\left( t\right) }^{n}\right| +1\right) \right) _{n=1}^{\infty }\right\| \right. \\&\quad +\left. \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( \frac{t}{n^{2}}\right) _{n=1}^{\infty }\right\| \right) \\\le & {} \frac{1}{2\left( t^{2}+1\right) }\left( \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( \left| y_{k\left( t\right) }^{n}\right| \right) _{n=1}^{\infty }\right\| \right. \\&\quad +\left. \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( \frac{t}{n^{2}}\right) _{n=1}^{\infty }\right\| \right) \\\le & {} \frac{1}{2}\left( \lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) x\right\| +\lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) \left( t\right) \right\| \right) \\\le & {} \frac{1}{2}\lim _{n\rightarrow \infty }\sup _{x\in B}\left\| \left( I-P_{n}\right) x\right\| +\frac{1}{2} \\= & {} \phi \left( \gamma \left( B\right) \right) , \end{aligned}$$

where \(L\left( t\right) =\left\| L\right\| =1\) and \(\phi \left( t\right) =\left( t+1\right) /2.\)

Also, it is easy to show that the other assumptions of Theorem 3.1 hold. Therefore, Theorem 3.1 guarantees that Eq. (3.10) has at least one solution \(x=x(t)\in B_{r_{0}}\subset C\left( J,c_{0}\right) \), such that \(r_{0}\ge 0.00640965\).