The class of functional integral equations of various types plays very important role in numerous mathematical research areas. An interesting feature of functional integral equations is its role in the study of many problems of functional differential Equations [1-4].
In this work we study the solvability of the following initial value problem
where 
denotes the fractional derivative of order 
of 
with
. Such initial value problem of arbitrary order (1) was investigated in [5-7]. To achieve this goal, let us consider the integral equation
which is different from that studied in [
Section 2 contains some basic results. Our main result will be given in Section 3. Solvability of the considered initial value problem will be discussed in Section 4.
is a Lebesgue measurable subset of the set of real numbers 
then we use the symbol 
to denote the Lebesgue measure of
. Let 
be the space of all real functions defined and Lebesgue measurable on the set
. If 
then the norm of 
is defined as:
we will write 
instead of 
satisfies Carathéodory conditions, that is it is measurable in 
for any 
and continuous in 
for almost all 
where
. Then for every measurable function 
on the interval 
we assign the function:
.
.
generated by 
will map continuously 
into itself [
satisfy the conditions in Definition 1. The superposition operator 
generated by the function 
maps continuously the space 
into itself if and only if:
and
, where 
is a function that belongs to 
and 
is a nonnegative constant.
be a bounded interval and 
be a function satisfying Caratheodory conditions. Then for each 
there exists a closed subset 
of the interval I such that 
and 
is continuous.
Suppose 
is a given function which is measurable with respect to both variables where
define Volterra integral operator as follows:
then it is continuous [4,9].
guaranteeing that the integral operator 
transforms the space 
into itself. Some special cases of this problem were discussed in [4,14]. In this direction we state the next result [
be measurable on 
and such that
generated by 
maps (continuously) the space 
into itself and the norm 
of this operator is majorized by the number
.
is a nonempty and measurable subset of 
then we can also consider the linear Volterra integral operator associated with the Lebesgue space 
where 
is a nonempty and measurable subset of 
then we extend 
to the whole half axis 
by putting 
for
. Then we can treat 
in the usual way. When the operator 
transforms 
into itself its norm will be denoted by
.
is an infinite dimensional Banach space with the norm 
and the zero element
. Denote by 
the family of all nonempty and bounded subsets of 
and by 
its subfamily consisting of all relatively weakly compact sets. The symbol 
stands for the weak closure of a set 
and the symbol 
will denote the convex closed hull (with respect to the norm topology) of a set
. We denote by 
the ball centered at 
and of radius
. We write 
instead of 
In what follows we accept the following definition [
is said to be a measure of weak noncompactness if it satisfies the following conditions: The Family 1) The family 
is nonempty and is nonempty and ker 
for 
.
then the intersection is nonempty 
is said to be the kernel of the measure of weak noncompactness
. Let us observe that the intersection set 
from 5) belongs to
. Indeed, since 
for every 
then we have that
.
that based on the following criterion for weak noncompactness due to Dieudonné [17,18].
is relatively weakly compact in 
if and only if the following two conditions are satisfied:
there exists 
such that if meas. 
Then 
for all,
.
there is 
such that 
for any ,
.
of the space 
let us define:
is a measure of weak noncompactness in the space 
such that
, for any
, where 
denotes the De Blasi measure of weak noncompactness in
. Moreover,
.
be a nonempty, convex, closed, and bounded subset of a Banach space
. Let 
be a completely continuous mapping. Then 
has at least one fixed point in
.
The RiemmanLiouville (R-L) fractional integral of the function 
of order 
is defined as
be an absolutely continuous function on
. Then the fractional derivative of order 
of 
is defined as
, then i) 
maps 
into itself continuously.
denotes the operator determined by the right hand side of this equation, i.e.,
In fact the operator 
can be written as the product 
of the linear Volterra operator
satisfy the Caratheodory conditions and there exist functions 
and constants 
such that
satisfy the Caratheodory conditions and the linear Volterra operators 
associated with 
map 
into itself.
is increasing, absolutely continuous and there exists a constant 
such that 
a.e. on
.
.
is a nonlinear operator defined by Equation (5), then based on assumptions i) and ii) if
, then 
Moreover, from Equation (5), and noting that 
according to our assumptions are indeed bounded, we have
maps 
into itself, where
is continuous on the space
.
is a contraction, let 
be a nonempty subset of 
Fix 
and take a measurable subset 
such that
. Then for any
, we get
denotes the operator norm acting from the space 
into itself. Also in the above calculation we used the fact that 
for
. From the absolute continuity of the function 
and the obvious equality
.
we can deduce that
denotes the operator norm acting from the space 
into itself. Now according to the fact that the set consisting of one element is weakly compact, by using Theorem 6 and the formula
we get
. Clearly, according to assumption iv)
. Consider the sequence of sets
, where 
and so on. Obviously this sequence is decreasing i.e. 
for 
Moreover,
. Apart from this, all sets belonging to this sequence are closed and convex, so weakly closed. On the other hand in view of inequality (8) we have
). Moreover,
.
is relatively compact in the set
.
and fix arbitrarily a number
. Since 
is weakly compact, in view of Theorem 6 we deduce that there exists 
such that for any natural number 
the following inequality is satisfied
is relatively compact in
. For this aim let us consider the functions 
on the set 
and the functions 
on the set
.
of the interval 
such that 
(where
) and such that the functions 
are continuous. Hence we infer that 
are uniformly continuous.
is an equicontinuous on
, for that let us take arbitrarily
. Without loss of generality we can assume that
. Then, keeping in mind our assumptions, for an arbitrary fixed 
we obtain:
denotes the modulus of continuity of the function 
on the set 
and
we obtain
is weakly compact and taking into account Theorem 6 we can show that the number
is taken to be sufficiently small (it is a consequence of the fact that a one element set is weakly compact in
).
is a sequence of uniformly bounded and equicontinuous functions on
. Hence, in view of Ascoli-Arzela theorem we deduce that the sequence 
is relatively compact subset in the space
.
. Thus we can construct a sequence 
of closed subsets of the interval 
such that 
as 
and such that the sequence 
is relatively compact in every space
. Passing to subsequences if necessary we can assume that 
is a Cauchy sequence in each space
, for 
is weakly compact, let us choose a number 
such that for each closed subset 
of the interval 
such that 
we have
.
is a Cauchy sequence in each space 
we can choose a natural number 
such that 
the following inequality holds
. Obviously without loss of generality we can assume that
.
is a Cauchy sequence in the space 
Hence we conclude that the set 
is relativelycompact in this space.
In view of the Mazur theorem we infer that the set 
is compact in the space
. Moreover, we have that the operator 
transforms continuously the set 
into itself. Thus the classical Schauder fixed point principle gives that 
has at least one fixed point. This proves that there exists at least one 
that solves Equation (4).
is an integrable function. For an arbitrary function 
set
is a linear integral operator of convolution type and maps 
into itself continuously.
.
. Equation (13) is an integral equation of fractional order that can be written in the form
.
and
.
.
be a solution of the integral Equation (14). Putting
is integrable, then
. Moreover, the integral 
of integrable function
is absolutely continuous then
is integrable and absolutely continuous, then
. Hence we deduce that 
is an absolutely continuous function satisfies the initial value problem (1). Hence the proof is complete.
is presented and proved. As an application of this theorem, we investigated the existence of solution of the suggested initial value problems of fractional order.