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Nonlinear fractional differential equations in nonreflexive Banach spaces and fractional calculus
Advances in Difference Equations volume 2015, Article number: 112 (2015)
Abstract
The aim of this paper is to correct some ambiguities and inaccuracies in Agarwal et al. (Commun. Nonlinear Sci. Numer. Simul. 20(1):5973, 2015; Adv. Differ. Equ. 2013:302, 2013, doi:10.1186/168718472013302) and to present new ideas and approaches for fractional calculus and fractional differential equations in nonreflexive Banach spaces.
1 Introduction
One of the sections, Section 5, of our paper [1] contains a number of ambiguities (and inaccuracies) which we correct here. The notion of pseudosolution in [1] is not adequately defined and assumption (h2) in Theorem 5.1 is strong. Also, there is some ambiguity regarding the use of the space \(C(T,E)\) of all continuous functions from T into E with its weak topology \(\sigma(C(T,E),C(T,E)^{\ast})\) and the space \(C_{w}(T,E)\) of all weakly continuous functions from T into \(E_{w}\) endowed with the topology of weak uniform convergence. Parts of Corollaries 5.15.6 are no longer valid in their current form. Similar comments also apply to [2]. In [3] the authors developed fractional calculus for vectorvalued functions using the weak Riemann integral and they established the existence of weak solutions for a class of fractional differential equations with fractional weak derivatives. In this paper we present new ideas in fractional calculus and we present a new approach to establishing existence to some fractional differential equations in nonreflexive Banach spaces. References [4–6], and [7] were helpful in presenting these new ideas.
2 Preliminaries
In the following we outline some aspects of fractional calculus in a nonreflexive Banach space. This subject has been treated extensively in [1, 3]. Let E be a Banach space with norm \(\Vert \cdot \Vert \) and let \(E^{\ast}\) be the topological dual of E. If \(x^{\ast}\in E^{\ast}\), then its value on an element \(x\in E\) will be denoted by \(\langle x^{\ast},x \rangle\). The space E endowed with the weak topology \(\sigma ( E,E^{\ast} ) \) will be denoted by \(E_{w}\). Consider an interval \(T= [ 0,b ] \) of ℝ, the set of real numbers, endowed with the Lebesgue σalgebra \(\mathcal{L} ( T ) \) and the Lebesgue measure λ. A function \(x(\cdot):T\rightarrow E\) is said to be strongly measurable on T if there exists a sequence of simple functions \(x_{n}(\cdot):T\rightarrow E\) such that \(\lim_{n\rightarrow\infty}x_{n}(t)=x(t)\) for a.e. \(t\in T\). Also, a function \(x(\cdot):T\rightarrow E\) is said to be weakly measurable (or scalarly measurable) on T if, for every \(x^{\ast}\in E^{\ast}\), the real valued function \(t\mapsto \langle x^{\ast },x(t) \rangle\) is Lebesgue measurable on T.
We denote by \(L^{p}(T)\) the space of all real measurable functions \(f:T\rightarrow\mathbb{R}\), whose absolute value raised to the pth power has finite integral, or equivalently, that
where \(1\leq p<\infty\). Moreover, by \(L^{\infty}(T)\) we denote the space of all measurable and essential bounded real functions defined on T. Let \(C(T,E)\) denote the space of all strong continuous functions \(y(\cdot ):T\rightarrow E\), endowed with the supremum norm \(\y(\cdot )\_{c}=\sup_{t\in T}\Vert y(t)\Vert \). Also, we consider the space \(C(T,E)\) with its weak topology \(\sigma(C(T,E),C(T,E)^{\ast})\). It is well known that (see [8, 9])
where \(M(T, E^{\ast})\) is the space of all bounded regular vector measures from \(\mathcal{B}(T)\) into \(E^{\ast}\) which are of bounded variation. Here, \(\mathcal{B}(T)\) denotes the σalgebra of Borel measurable subsets of T. Therefore, a sequence \(\{y_{n}(\cdot)\}_{n\geq1}\) converges weakly to \(y(\cdot)\) in \(C(T,E)\) if and only if
for all \(m(\cdot)\in M(T,E^{\ast})\). In [10], Lemma 9, it is shown that a sequence \(\{y_{n}(\cdot)\}_{n\geq1}\) converges weakly to \(y(\cdot)\) in \(C(T_{0},E)\) if and only if \(y_{n}(t)\) tends weakly to \(y(t)\) for each \(t\in T\).
Let \(C_{w}(T,E)\) denote the space of all weakly continuous functions from T into \(E_{w}\) endowed with the topology of weak uniform convergence. A set \(N\in\mathcal{L} ( T ) \) is called a null set if \(\lambda ( N ) =0\).
A function \(x(\cdot):T\rightarrow E\) is said to be pseudodifferentiable on T to a function \(y(\cdot):T\rightarrow E\) if, for every \(x^{\ast}\in E^{\ast}\), there exists a null set \(N(x^{\ast })\in \mathcal{L} ( T ) \) such that the real function \(t\mapsto \langle x^{\ast},x(t) \rangle\) is differentiable on \(T\smallsetminus N(x^{\ast})\) and
The function \(y(\cdot)\) is called a pseudoderivative of \(x(\cdot)\) and it will be denoted by \(x_{p}^{\prime}(\cdot)\) or by \(\frac {d_{p}}{dt}x(\cdot)\). A pseudoderivative \(x_{p}^{\prime}(\cdot)\) of a pseudodifferentiable function \(x(\cdot):T\rightarrow E\) is weakly measurable on T (see [11]).
We recall that a function \(x(\cdot):T\rightarrow E\) is said to be weakly differentiable on T if there exists a function \(x_{w}^{\prime}(\cdot):T\rightarrow E\) such that
for every \(x^{\ast}\in E^{\ast}\). If it exists, \(x_{w}^{\prime }(\cdot)\) is uniquely determined and it is called the weak derivative of \(x(\cdot)\) on T. Obviously, if \(x(\cdot):T\rightarrow E\) is a weakly differentiable function on T, then the real function \(t\mapsto \langle x^{\ast},x(t) \rangle\) is differentiable on T. Moreover, in this case we have
for every \(x^{\ast}\in E^{\ast}\). It is easy to see that, if \(x(\cdot ):T\rightarrow E\) is a function a.e. weakly differentiable on T, then \(x(\cdot)\) is pseudodifferentiable on T and \(x_{p}^{\prime}(\cdot )=x_{w}^{\prime}(\cdot)\) a.e. on T.
The concept of a Bochner integral and a Pettis integral are well known [12–14].
We recall that a weakly measurable function \(x(\cdot ):T\rightarrow E\) is said to be Pettis integrable on T if

(a)
\(x(\cdot)\) is scalarly integrable; that is, for every \(x^{\ast}\in E^{\ast}\), the real function \(t\mapsto \langle x^{\ast },x(t) \rangle\) is Lebesgue integrable on T;

(b)
for every set \(A\in\mathcal{L} ( T ) \), there exists an element \(x_{A}\in E\) such that
$$ \bigl\langle x^{\ast},x_{A} \bigr\rangle =\int _{A} \bigl\langle x^{\ast },x(s) \bigr\rangle \,ds, $$(3)for every \(x^{\ast}\in E^{\ast}\). The element \(x_{A}\in E\) is called the Pettis integral on A and it will be denoted by \(\int_{A}x(s)\,ds\).
It is easy to show that a Bochner integrable function \(x(\cdot ):T\rightarrow E\) is Pettis integrable and both integrals of \(x(\cdot )\) are equal on each Lebesgue measurable subset A of T ([14], Proposition 2.3.1). The best result for a descriptive definition of the Pettis integral is that given by Pettis in [15].
Proposition 1
Let \(x(\cdot):T\rightarrow E\) be a weakly measurable function.

(a)
If \(x(\cdot)\) is Pettis integrable on T, then the indefinite Pettis integral
$$ y(t):=\int_{0}^{t}x(s)\,ds,\quad t\in T $$is AC on T and \(x(\cdot)\) is a pseudoderivative of \(y(\cdot)\).

(b)
If \(y(\cdot):T\rightarrow E\) is an AC function on T and it has a pseudoderivative \(x(\cdot )\) on T, then \(x(\cdot)\) is Pettis integrable on T and
$$ y(t)=y(0)+\int_{0}^{t}x(s)\,ds,\quad t\in T. $$
It is well known that the Pettis integrals of two strongly measurable functions \(x(\cdot):T\rightarrow E\) and \(y(\cdot):T\rightarrow E\) coincide over every Lebesgue measurable set in T if and only if \(x(\cdot )=y(\cdot)\) a.e. on T ([15], Theorem 5.2). Since a pseudoderivative of a pseudodifferentiable function \(x(\cdot ):T\rightarrow E\) is not unique (see [11]) and two pseudoderivatives of \(x(\cdot) \) need not be a.e. equal, the concept of weakly equivalence plays an important role in the following.
Two weak measurable functions \(x(\cdot):T\rightarrow E\) and \(y(\cdot):T\rightarrow E\) are said to be weakly equivalent on T if, for every \(x^{\ast}\in E^{\ast}\), we have \(\langle x^{\ast },x(t) \rangle= \langle x^{\ast},y(t) \rangle\) for a.e. \(t\in T\). In the following, if two weak measurable functions \(x(\cdot ):T\rightarrow E\) and \(y(\cdot):T\rightarrow E\) are weakly equivalent on T, then we will write \(x(\cdot)\eqsim y(\cdot)\) or \(x(t)\eqsim y(t)\), \(t\in T\).
Proposition 2
([15])
A weakly measurable function \(x(\cdot):T\rightarrow E\) is Pettis integrable on T and \(\langle x^{\ast},x(\cdot) \rangle\in L^{\infty }(T)\), for every \(x^{\ast}\in E^{\ast}\), if and only if the function \(t\mapsto\varphi(t)x(t)\) is Pettis integrable on T, for every \(\varphi(\cdot)\in L^{1}(T)\).
Let us denote by \(P^{\infty}(T,E)\) the space of all weakly measurable and Pettis integrable functions \(x(\cdot):T\rightarrow E\) with the property that \(\langle x^{\ast},x(\cdot) \rangle\in L^{\infty}(T)\), for every \(x^{\ast}\in E^{\ast}\). Since for each \(t\in T\) the real valued function \(s\mapsto(ts)^{\alpha1}\) is Lebesgue integrable on \([ 0,t ] \), the fractional Pettis integral
exists, for every function \(x(\cdot)\in P^{\infty}(T,E)\), as a function from T into E (see [16]). Moreover, we have
for every \(x^{\ast}\in E^{\ast}\), and the real function \(t\mapsto \langle x^{\ast},I^{\alpha}x(t) \rangle\) is continuous (in fact, bounded and uniformly continuous on T if \(T=\mathbb{R}\)) on T, for every \(x^{\ast}\in E^{\ast}\) ([17], Proposition 1.3.2).
In the following, consider \(\alpha\in(0,1)\) and for a given function \(x(\cdot)\in P^{\infty}(T,E)\) we also denote by \(x_{1\alpha }(t)\) the fractional Pettis integral
Lemma 1
([1], Lemma 3.1)
If \(x(\cdot ),y(\cdot)\in P^{\infty}(T,E)\) are weakly equivalent on T, then \(I^{\alpha}x(t)=I^{\alpha}y(t)\) on T.
Lemma 2
The fractional Pettis integral is a linear operator from \(P^{\infty}(T,E)\) into \(P^{\infty}(T,E)\). Moreover, if \(x(\cdot)\in P^{\infty }(T,E)\), then for \(\alpha,\beta>0\) we have

(a)
\(I^{\alpha}I^{\beta}x(t)=I^{\alpha+\beta}x(t)\), \(t\in T\);

(b)
\(\lim_{\alpha\rightarrow1}I^{\alpha }x(t)=I^{1}x(t)=x(t)x(0)\) weakly uniformly on T;

(c)
\(\lim_{\alpha\rightarrow0}I^{\alpha}x(t)=x(t)\) weakly on T.
If \(y(\cdot):T\rightarrow E\) is a pseudodifferentiable function on T with a pseudoderivative \(x(\cdot)\in P^{\infty}(T,E)\), then the fractional Pettis integral \(I^{1\alpha}x(t)\) exists on T. The fractional Pettis integral \(I^{1\alpha}x(\cdot)\) is called a fractional pseudoderivative of \(y(\cdot)\) on T and it will be denoted by \(D_{p}^{\alpha}y(\cdot)\); that is,
Remark 1
If \(x(\cdot),\widetilde{x}(\cdot)\in P^{\infty}(T,E)\) are two pseudoderivatives of \(y(\cdot):T\rightarrow E\), then \(x(\cdot)\eqsim\widetilde{x}(\cdot)\) on T. Thus, Lemma 1 implies that \(I^{1\alpha}x(t)=I^{1\alpha}\widetilde{x}(t)\) on T, and so \(D_{p}^{\alpha}y(\cdot)\) does not depend on the choice of a pseudoderivatives of the function \(y(\cdot)\). Therefore, we can write (4) as
where \(y_{p}^{\prime}(\cdot)\) is a given pseudoderivatives of \(y(\cdot)\).
We recall that a function \(x(\cdot):T\rightarrow E\) is said to be weakly absolutely continuous (\(wAC\), for short) on T if, for every \(x^{\ast}\in E^{\ast}\), the real valued function \(t\mapsto \langle x^{\ast},x(t) \rangle\) is absolutely continuous on T.
Lemma 3
([1])
If \(y(\cdot)\in P^{\infty }(T,E)\) is a pseudodifferentiable function on T with a pseudoderivative \(x(\cdot)\in P^{\infty}(T,E)\), then the function
is \(wAC\) and it has a pseudoderivative \(\frac {d_{p}}{dt}y_{1\alpha}(\cdot)\in P^{\infty}(T,E)\) such that
Remark 2
Relation (6) can be written as
Note (7) suggests us that we can extend the definition of the fractional pseudoderivative for functions \(y(\cdot)\in P^{\infty }(T,E) \) for which the function \(t\mapsto y_{1\alpha}(t)\) is pseudodifferentiable on T. If \(\frac{d_{p}}{dt}y_{1\alpha}(t)\) exists on T, then \(\frac{d_{p}}{dt}y_{1\alpha}(t)\) will be called the RiemannLiouville fractional pseudoderivative of \(y(\cdot)\) and it will be denoted by \(\mathcal{D}_{p}^{\alpha}y(\cdot)\); that is, \(\mathcal {D}_{p}^{\alpha}y(\cdot)=\frac{d_{p}}{dt}y_{1\alpha}(\cdot)\). Usually, \(D_{p}^{\alpha}y(\cdot)\) is called the Caputo fractional pseudoderivative of \(y(\cdot)\). Relation (6) can be written as
Therefore, the Caputo fractional pseudoderivative \(D_{p}^{\alpha }y(\cdot)\) exists together with the RiemannLiouville fractional pseudoderivative \(\mathcal{D}_{p}^{\alpha}y(\cdot)\) and they satisfy (8). It is easy to see that if \(y(0)=0\), then
Remark 3
Let \(y(\cdot):T\rightarrow E\) be a pseudodifferentiable function with a pseudoderivative \(y_{p}^{\prime }(\cdot)\in P^{\infty}(T,E)\). Then from Lemma 3 we find that the function
is \(wAC\) and has a pseudoderivative \(\frac{d_{p}}{dt}y_{\alpha}(t)\) such that
Lemma 4
Let \(\alpha,\beta\in(0,1)\).

(a)
If \(y(\cdot)\in P^{\infty}(T,E)\), then
$$ D_{p}^{\alpha}I^{\alpha}y(t)=y(t),\quad t\in T. $$ 
(b)
If \(y(\cdot)\in P^{\infty}(T,E)\) and \(y_{1\alpha }(\cdot)\) is pseudodifferentiable with a pseudoderivative \(\frac{d_{p}}{dt}y_{1\alpha}(\cdot)\in P^{\infty}(T,E)\), then
$$ I^{\alpha}D_{p}^{\alpha}y(t)=y(t)y(0),\quad t\in T. $$
Proof
(a) Indeed, since \(y(\cdot)\in P^{\infty}(T,E)\), then \(t\mapsto \langle x^{\ast},y(t) \rangle\) is essentially bounded on T, for every \(x^{\ast}\in E^{\ast}\). Hence we have
where \(M(x^{\ast})=\operatorname{ess\,sup}_{t\in T}\vert \langle x^{\ast},y(t) \rangle \vert <\infty\), \(x^{\ast}\in E^{\ast}\). Since the real function \(t\mapsto \langle x^{\ast},I^{\alpha }y(t) \rangle\) is continuous on T, it follows that \(\langle x^{\ast},I^{\alpha}y(0) \rangle=0\), for every \(x^{\ast}\in E^{\ast}\), and thus \(I^{\alpha}y(0)=0\). Then by Remark 2 we have \(D_{p}^{\beta }I^{\alpha}y(t)=\mathcal{D}_{p}^{\beta}I^{\alpha}y(t)\), and so by Lemma 2 and Proposition 1 we have
(b) By Lemma 2 and Proposition 1 we have
□
Lemma 5
Let \(y(\cdot):T\rightarrow E\) be a pseudodifferentiable function on T with \(y_{p}^{\prime }(\cdot)\in P^{\infty}(T,E)\) and \(0<\alpha\leq\beta <1\). Then we have

(a)
$$ I^{\alpha}D_{p}^{\beta}y(t)=D_{p}^{\beta\alpha}y(t) \quad\textit{on } T. $$(10)

(b)
If \(y(0)=0\), then
$$ D_{p}^{\beta}I^{\alpha}y(t)=D_{p}^{\beta\alpha}y(t) \quad\textit{on }T $$(11)and
$$ I^{\beta}D_{p}^{\alpha}y(t)=I^{\beta\alpha}y(t)\quad \textit{on }T. $$(12)
Proof
If \(y(\cdot):T\rightarrow E\) is a pseudodifferentiable function on T, then by Lemma 2 we have
If \(y(0)=0\), then by Remark 3 and (10) we have
Also, since \(y(0)=0\), then by Lemma 2 and Proposition 1 we have
□
3 Differential equations with fractional pseudoderivatives
The existence of weak solutions or pseudosolutions for ordinary differential equations in Banach spaces were investigated in many papers (see [18–31]). In reflexive Banach spaces, the existence of weak solutions or pseudosolutions for fractional differential equations were studied in [32–36]. In this section we establish an existence result for the following fractional differential equation:
where \(D_{p}^{\alpha}y(\cdot)\) is a fractional pseudoderivative of the function \(y(\cdot):T\rightarrow E\) and \(f(\cdot,\cdot):T\times E\rightarrow E\) is a given function. Along with the Cauchy problem (13) consider the following integral equation:
where the integral is in the sense of Pettis.
A continuous function \(y(\cdot):T\rightarrow E\) is said to be a solution of (13) if \(y(\cdot)\) has a pseudoderivative belonging to \(P^{\infty}(T,E)\), \(D_{p}^{\alpha}y(t)\eqsim f(t, y(t))\) for \(t\in T\) and \(y(0)=y_{0}\).
To prove a result on the existence of solutions for (13) we need some preliminary results.
Lemma 6
Let \(f(\cdot,\cdot):T\times E\rightarrow E\) be a function such that \(f(\cdot,y(\cdot))\in P^{\infty}(T,E)\), for every continuous function \(y(\cdot ):T\rightarrow E\). Then a continuous function \(y(\cdot ):T\rightarrow E\) is a solution of (13) if and only if it satisfies the integral equation (14).
Proof
Indeed, if a continuous function \(y(\cdot):T\rightarrow E\) is a solution of (13), then \(y(\cdot )\) has a pseudoderivative belonging to \(P^{\infty}(T,E)\), \(D_{p}^{\alpha }y(t)\eqsim f(t, y(t))\) for \(t\in T\) and \(y(0)=y_{0}\). Then we have \(I^{\alpha}D_{p}^{\alpha}y(t)=I^{\alpha}f(t,y(t))\) on T, and thus from Lemma 4(b) it follows that \(y(t)y(0)=I^{\alpha }f(t,y(t))\) on T; that is, \(y(\cdot)\) satisfies the integral equation (14). Conversely, suppose that a continuous function \(y(\cdot ):T\rightarrow E\) satisfies the integral equation (14). Then the function \(z(\cdot):=f(\cdot,y(\cdot))\in P^{\infty}(T,E) \) satisfies the Abel equation
where \(v(t):=y(t)y_{0}\), \(t\in T\). Then from [1], Theorem 3.1, and Lemma 3 it follows that \(v_{1\alpha}(\cdot)\) has a pseudoderivative on T and
Then by Remark 2 we have \(z(t)\eqsim D_{p}^{\alpha}y(t) \) for \(t\in T\); that is, \(D_{p}^{\alpha}y(t)\eqsim f(t, y(t))\) on T. □
In this section we shall discuss the existence of solutions of fractional differential equations in nonreflexive Banach spaces. We recall that a function \(f(\cdot):E\rightarrow E\) is said to be sequentially continuous from \(E_{w}\) into \(E_{w}\) (or weaklyweakly sequentially continuous) if, for every weakly convergent sequence \(\{x_{n}\}_{n\geq1}\subset E\), the sequence \(\{f(x_{n})\}_{n\geq1}\) is weakly convergent in E.
By a Gripenberg function we mean a function \(g:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}\) such that \(g(\cdot)\) is continuous, nonincreasing with \(g(0)=0\), and \(u\equiv0\) is the only continuous solution of
The problem of uniqueness of the null solution of (15) was studied by Gripenberg in [37].
Let us denote by \(P_{wk}(E)\) the set of all weakly compact subset of E. The weak measure of noncompactness [38] is the set function \(\beta:P_{wk}(E)\rightarrow[0,\infty)\) defined by
where \(B_{1}\) is the closed unit ball in E. The properties of the weak noncompactness measure are analogous to the properties of the measure of noncompactness, namely (see [38]):

(1)
\(A\subseteq B\) implies that \(\beta(A)\leq\beta(B)\);

(2)
\(\beta(A)=\beta(cl_{w}A)\), where \(cl_{w}A\) denotes the weak closure of A;

(3)
\(\beta(A)=0\) if and only if \(cl_{w}A\) is weakly compact;

(4)
\(\beta(A\cup B)=\max\{\beta(A),\beta(B)\}\);

(5)
\(\beta(A)=\beta(\operatorname{conv}(A))\);

(6)
\(\beta(A+B)\leq\beta(A)+\beta(B)\);

(7)
\(\beta(x+A)=\beta(A)\), for all \(x\in E\);

(8)
\(\beta(\lambda A)=\lambda\beta(A)\), for all \(\lambda\in \mathbb{R}\);

(9)
\(\beta(\bigcup_{0\leq r\leq r_{0}}rA)=r_{0}\beta(A)\);

(10)
\(\beta(A)\leq2\operatorname{diam}(A)\).
Lemma 7
([39])
Let \(H\subset C(T,E)\) be bounded and equicontinuous. Then

(i)
the function \(t\rightarrow\beta(H(t)) \) is continuous on T,

(ii)
\(\beta_{c}(H)=\sup_{t\in T}\beta(H(t))\),
where \(\beta_{c}(\cdot)\) denote the weak noncompactness measure on \(C(T,E)\) and \(H(t)=\{u(t), u\in H\}\), \(t\in T\).
Lemma 8
([21])
Let E be a metrizable locally convex topological vector space and let K be a closed convex subset of E, and let Q be a weakly sequentially continuous map of K into itself. If for some \(y\in K\) the implication
holds, for every subset V of K, then Q has a fixed point.
Theorem 1
Assume \(f(\cdot,\cdot):T\times E\rightarrow E\) is a function such that:

(H1)
\(f(t,\cdot)\) is weaklyweakly sequentially continuous, for every \(t\in T\);

(H2)
\(f(\cdot,y(\cdot))\in P^{\infty}(T,E)\), for every continuous function \(y(\cdot):T\rightarrow E\);

(H3)
\(\f(t,y)\\leq M\), for all \((t,y)\in T\times E\);

(H4)
for every bounded set \(A\subseteq E\) we have
$$ \beta\bigl(f(T\times A)\bigr)\leq g\bigl(\beta(A)\bigr), $$
where \(g(\cdot)\) is a Gripenberg function. Then (13) admits a solution \(y(\cdot)\) on an interval \(T_{0}=[0,a]\) with \(a=\min \{ b, ( \frac{\Gamma(\alpha +1)}{M} ) ^{1/\alpha} \} \).
Proof
In our proof we shall use some ideas from [5] and [6]. We define the nonlinear operator \(Q ( \cdot ) :C(T_{0},E)\rightarrow C(T_{0},E)\) by
If \(y(\cdot)\in C(T_{0},E)\), then by (H2) we have \(f(\cdot,y(\cdot ))\in P^{\infty}(T,E)\) and so the operator Q makes sense. To show that Q is well defined, let \(t_{1},t_{2}\in T_{0}\) with \(t_{2}>t_{1}\). Without loss of generality, assume that \((Qy)(t_{2})(Qy)(t_{1})\neq0\). Then by the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\) and \(\Vert (Qy)(t_{2})(Qy)(t_{1})\Vert =\vert \langle y^{\ast },(Qy)(t_{2})(Qy)(t_{1}) \rangle \vert \). Then
so Q maps \(C(T_{0},E)\) into itself. Let K be the convex, closed, and equicontinuous set defined by
We will show that Q maps K into itself and Q restricted to the set K is weaklyweakly sequentially continuous. To show that \(Q:K\rightarrow K\), let \(y(\cdot)\in K\) and \(t\in T_{0}\). Again, without loss of generality, assume that \((Qy)(t)\neq0\). By the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\) and \(\ (Qy)(t)\=\vert \langle y^{\ast},(Qy)(t) \rangle \vert \). Then by (H3), we have
and using (16) it follows that Q maps K into K. Next, we show that Q is weaklyweakly sequentially continuous. First, we recall that the weak convergence in \(K\subset C(T_{0},E)\) is exactly the weak pointwise convergence. Let \(\{y_{n}(\cdot)\}_{n\geq1}\) be a sequence in K such that \(y_{n}(\cdot)\) converges weakly to \(y(\cdot)\) in K. Then \(y_{n}(t)\) converges weakly to \(y(t)\) in E for each \(t\in T_{0}\). Since K is a closed convex set, by Mazur’s lemma we have \(y(\cdot)\in K\). Further, by (H1) it follows that \(f(t,y_{n}(t))\) converges weakly to \(f(t,y(t))\) for each \(t\in T_{0}\). Then the Lebesgue dominated convergence theorem for the Pettis integral (see [40]) yields \(I^{\alpha}y_{n}(t)\) converging weakly to \(I^{\alpha}y(t)\) in E for each \(t\in T_{0}\). Since K is an equicontinuous subset of \(C(T_{0},E)\) it follows that \(Q(\cdot)\) is weaklyweakly sequentially continuous.
Suppose that \(V\subset K\) is such that \(V=\overline {\operatorname{co}}(Q(V)\cup \{y(\cdot)\})\) for some \(y(\cdot)\in K\). We will show that V is relatively weakly compact in \(C(T_{0},E)\). Let
and \((QV)(t)=y_{0}+\int_{0}^{t}\frac{(ts)^{\alpha1}}{\Gamma(\alpha )}f(s,V(s))\,ds\). Let \(t\in T_{0}\) and \(\varepsilon>0\). If we choose \(\eta>0\) such that \(\eta< ( \frac{\varepsilon\Gamma(\alpha+1)}{M} ) ^{1/\alpha}\) and \(\int_{t\eta}^{t}\frac{(ts)^{\alpha1}}{\Gamma (\alpha )}f(s,y(s))\,ds\neq0\) then, by the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\) and
It follows that
and thus using property (10) of the noncompactness measure we infer
Since by Lemma 7 the function \(s\rightarrow v(s):=\beta(V(s))\) is continuous on \([0,t\eta]\) it follows that \(s\rightarrow(ts)^{\alpha 1}g(v(s))\) is continuous on \([0,t\eta]\). Hence, there exists \(\delta>0\) such that
and
for all \(\tau,s,\xi\in[0,t\eta]\) with \(\taus<\delta\) and \(\tau\xi<\delta\). It follows that
that is,
for all \(\tau,s,\xi\in[0,t\xi] \) with \(\taus<\delta\) and \(\tau\xi<\delta\). Consider a partition of the interval \([0,t\eta]\) into n parts \(0=t_{0}< t_{1}<\cdots<t_{n}=t\eta\) such that \(t_{i}t_{i1}<\delta\), \(i=1,2,\ldots,n\). From Lemma 7 it follows that for each \(i\in\{1,2,\ldots,n\}\) there exists \(s_{i}\in[ t_{i1},t_{i}]\) such that \(\beta(V([t_{i1},t_{i}]))=v(s_{i})\), \(i=1,2,\ldots,n\). Then we have (see [41], Theorem 2.2)
and so
Using (18) we have
This implies that
Thus we obtain
Now because
then from (17) and (19) we have
As the last inequality is true, for every \(\varepsilon>0\), we infer
Because \(V=\overline{\operatorname{co}}(Q(V)\cup\{y(\cdot)\})\) then
and thus
Since \(g(\cdot)\) is a Gripenberg function, it follows that \(v(t)=0\) for \(t\in T_{0}\). Since V as a subset of K is equicontinuous, by Lemma 7 we infer
Thus, by ArzeláAscoli’s theorem we find that V is weakly relatively compact in \(C(T_{0},E)\). Using Lemma 8 there exists a fixed point of the operator Q which is a solution of (13). □
If E is reflexive and \(f(\cdot,\cdot):T\times E\rightarrow E\) is bounded, then (H4) is automatically satisfied since a subset of a reflexive Banach space is weakly compact iff it is closed in the weak topology and bounded in the norm topology.
If for \(\alpha=1\) we put \(D_{p}^{1}y(\cdot)=y_{p}^{\prime}(\cdot)\), then from Theorem 1 we obtain the following result (see [18, 23]).
Corollary 1
If \(f(\cdot,\cdot):T\times E\rightarrow E\) is a function that satisfies the conditions (H1)(H4) in Theorem 1, then the differential equation
has a solution on \([0,a]\) with \(a=\min\{b,1/M\}\).
4 Multiterm fractional differential equation
The case of multiterm fractional differential equations in reflexive Banach spaces was recently considered in [42–45]. Consider the following multiterm fractional differential equation:
where \(D^{\alpha_{i}}y(\cdot)\), \(i=1,2,\ldots,m\), are fractional pseudoderivatives of order \(\alpha_{i}\in(0,1)\) of a pseudodifferentiable function \(y(\cdot):[0,1]\rightarrow E\), \(f(t,\cdot ):[0,1]\times E\rightarrow E\) is weaklyweakly sequentially continuous, for every \(t\in[0,1]\), and \(f(\cdot,y(\cdot))\) is Pettis integrable, for every continuous function \(y(\cdot):[0,1]\rightarrow E\), E is a nonreflexive Banach space, \(0<\alpha_{1}<\alpha_{2}<\cdots<\alpha_{m}<1\) and \(a_{1},a_{2},\ldots,a_{m1}\) are real numbers such that \(a:=\sum_{i=1}^{m1}\frac{a_{i}}{\Gamma(\alpha_{m}\alpha _{i}+1)}<1\).
Along with the Cauchy problem (21) consider the following integral equation:
\(t\in T\), where the integral is in the sense of Pettis and \(T=[0,1]\).
A continuous function \(y(\cdot):T\rightarrow E\) is said to be a solution of (21) if

(i)
\(y(\cdot)\) has Caputo fractional pseudoderivatives of orders \(\alpha _{i}\in(0,1)\), \(i=1,2,\ldots,m\),

(ii)
\(( D^{\alpha_{m}}\sum_{i=1}^{m1}a_{i}D^{\alpha _{i}} ) y(t)\eqsim f(t, y(t))\), for all \(t\in T\),

(iii)
\(y(0)=0\).
Lemma 9
Assume that \(f(\cdot,\cdot ):T\times E\rightarrow E\) satisfy the assumptions (H2) and (H3) in Theorem 1. Then every continuous function \(y(\cdot):T\rightarrow E\) which satisfies the integral equation (22) is a solution of (21).
Proof
Suppose that a continuous function \(y(\cdot ):T\rightarrow E\) satisfies the integral equation (14). Then \(z(\cdot):=f(\cdot,y(\cdot))\in P^{\infty}(T,E) \) satisfies the Abel equation
where \(v(t):=y(t)\sum_{i=1}^{m1}a_{i}I^{\alpha_{m}\alpha _{i}}y(t)\), \(t\in T\). From [1], Theorem 3.1, it follows that \(v_{1\alpha _{m}}(\cdot)\) has a pseudoderivative on T and
Since \(y(\cdot)\) is continuous on T and \(f(\cdot,y(\cdot))\in P^{\infty }(T,E)\) satisfies (H3), we have
for \(\alpha\in(0,1)\) and thus, taking the limit as \(t\rightarrow0^{+}\) on both equalities in (22), we obtain \(y(0)=0\) and consequently \(v(0)=0\). Since \(v(0)=0\), by Remark 2 we have
Since by Lemma 5(b) we have
we obtain
Hence the continuous function \(y(\cdot)\) satisfy the conditions (i)(iii) from definition and thus \(y(\cdot)\) is a solution of (21). □
Lemma 10
([24], Theorem 2.2)
Let K be a nonempty, bounded, convex, closed set in a Banach space E. Assume \(Q:K\rightarrow K\) is weakly sequentially continuous and βcontractive (that is, there exists \(0\leq k_{0}<1\) such that \(\beta(Q(A))\leq k_{0}\beta(A)\), for all bounded sets \(A\subset E\)). Then Q has a fixed point.
Remark 4
Since the function \(\sigma\mapsto\Gamma (\sigma)\) is convex and \(\Gamma(\sigma)\geq\Gamma(3/2)\approx0.8856032\) for \(\sigma\in(1,2)\), for every \(r\in(0,\Gamma(3/2))\) we have \(\Gamma(\alpha_{m}+1)>r\).
Next we establish an existence result for the multiterm fractional integral equation (22) in nonreflexive Banach spaces.
Theorem 2
Suppose that \(f(\cdot ,\cdot ):T\times E\rightarrow E\) satisfies the conditions (H1)(H3) in Theorem 1 and there exists \(L>0\) such that, for every bounded set \(A\subseteq E\), we have
If \(r\in(0,1)\) is such that \(\Gamma(\alpha _{m}+1)>r\), then (22) admits a solution \(y(\cdot)\) on an interval \(T_{0}=[0,a_{0}]\) with
Proof
We define the nonlinear operator \(Q ( \cdot ) :C(T_{0},E)\rightarrow C(T_{0},E)\) by
for all \(t\in T_{0}\). We remark that a solution of integral equation ( 22 ) is a fixed point of the operator Q. If \(y(\cdot)\in C(T_{0},E)\), then by (H2) we have \(f(\cdot,y(\cdot))\in P^{\infty}(T_{0},E)\) and so the operator Q makes sense. To show that Q is well defined, let \(t,s\in T_{0}\) with \(t>s\). Without loss of generality, assume that \((Qy)(t)(Qy)(s)\neq0\). Then by the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\) and \(\Vert (Qy)(t)(Qy)(s)\Vert =\vert \langle y^{\ast },(Qy)(t)(Qy)(s) \rangle \vert \). Then
so Q maps \(C(T_{0},E)\) into itself. Let \(\delta\geq1\) and let K be the convex, closed, bounded and equicontinuous set defined by
Without loss of generality, assume that \((Qy)(t)\neq0\). By the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast }\Vert =1\) and \(\(Qy)(t)\=\vert \langle y^{\ast },(Qy)(t) \rangle \vert \). Then by (H3), we have
and using (23) it follows that Q maps K into K. Following the same reasoning as in the proof of Theorem 1 it is easy to show that Q is weaklyweakly sequentially continuous from K to K. Next, we will prove that Q has at least one fixed point \(y_{0}(\cdot)\in K\). Let \(V\subset K\) be such that \(\beta_{c}(V)>0\). Next, to simplify the writing of some relations, we will use the following notations:
for \(t\in T_{0}\). Then it is easy to see that \(\vert \langle y^{\ast},C(t) \rangle \vert \leq\frac{M\eta^{\alpha _{m}}}{\Gamma(\alpha_{m}+1)}\) and \(\vert \langle y^{\ast },D(t) \rangle \vert \leq\sum_{i=1}^{m1}\frac{ra_{i}\eta ^{\alpha_{{m}}\alpha_{i}}}{\Gamma(\alpha_{m}\alpha_{i}+1)}\), for all \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\). Let \(t\in T_{0}\) and \(\varepsilon>0\). If we choose \(\eta>0\) such that \(\eta < ( \frac{\varepsilon\Gamma(\alpha_{m}+1)}{r [ M+\Gamma(\alpha _{m}+1) ] } ) ^{1/\alpha_{m}}\) and \(C(t)+D(t)\neq0\), then by the HahnBanach theorem, there exists a \(y^{\ast}\in E^{\ast}\) with \(\Vert y^{\ast} \Vert =1\) and
and thus using property (10) of the measure of noncompactness we infer
As in the proof of Theorem 1, with \(g(t)=L\), \(t\in T_{0}\), we obtain
Also, with \(y(\cdot)\) instead of \(f(\cdot,y(\cdot))\), we have
and so
Next, since \((QV)(t)=(AV)(t)+(CV)(t)+(BV)(t)+(DV)(t)\), \(t\in T_{0}\), then from the last inequalities and using properties of the noncompactness measure we infer
As the last inequality is true, for every \(\varepsilon>0\), it follows that
Since \(\beta(V(t))\leq\beta_{c}(V)\), \(t\in T_{0}\), we have
where \(k_{0}=a_{0} ( 1+\frac{L}{r} ) <1\). It follows that \(\beta _{c}(QV)< k_{0}\beta_{c}(V)\), for every set \(V\subset K\) with \(\beta _{c}(V)>0 \); that is, \(Q:K\rightarrow K\) is a \(\beta_{c}\)contractive operator. Since K is a nonempty, closed, convex, bounded subset in \(C(T_{0},E)\), and \(Q:K\rightarrow K\) is weakly sequentially continuous and \(\beta_{c}\)contractive, by Lemma 10 it follows that the operator Q has a fixed point \(y_{0}(\cdot)\in K\). □
Using Lemma 9 we obtain the following result.
Corollary 2
If the assumptions of Theorem 2 are satisfied, then the problem (21) has at least one solution.
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Acknowledgements
The authors wish to thank Professor Mieczysław Cichoń for his helpful comments.
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Agarwal, R.P., Lupulescu, V., O’Regan, D. et al. Nonlinear fractional differential equations in nonreflexive Banach spaces and fractional calculus. Adv Differ Equ 2015, 112 (2015). https://doi.org/10.1186/s1366201504515
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DOI: https://doi.org/10.1186/s1366201504515
Keywords
 Banach Space
 Weak Solution
 Fractional Calculus
 Fractional Differential Equation
 Weak Topology