We assume the following.

(A1)
If and a nonnegative, bounded , there exist , for such that

(A2)
There exist positive constants , , and such that

for
all , , and .

(A3)
There exist positive constants , , and such that

for
all , , and .

(A4)
There exist positive constants , , and such that

for
all , , and .

(A5)
is such that .
Firstly, we obtain the following lemmas to prove the main results on the existence of solutions to (1.2).
Lemma 3.1.
If (A1) holds with , then the problem (1.2) is equivalent to the following equation:
Proof.
By Lemma 2.5 and (1.2), we have
Therefore,
So,
and then
Conversely, if is a solution of (3.5), then for every , according to Definition 2.4 we have
It is obvious that . This completes the proof.
Lemma 3.2.
If (A3) and (A4) are satisfied, , are defined in (A), then the conditions
are satisfied for any , and .
Proof.
By (A3), we have
On the other hand,
Similarly, for the other conditions, we use assumption (A4), to get
Theorem 3.3.
If (A1)–(A5) are satisfied, then the fractional integrodifferential equation (1.2) has a unique solution continuous in .
Proof.
We use the Banach contraction principle to prove the existence and uniqueness of the solution to (1.2). Let , where and define the operator on the Banach space by
Firstly, we show that the operator maps into itself. By using (A1) and triangle inequality, we have
Now, if (A2) is satisfied, then
Using Lemma 3.2 and (A3), we have
if , we have
Thus . Next, we prove that is a contraction mapping. For this, let . Applying (A2), we have
then using (A3), (A4) and Lemma 3.2, one gets
where = depends on the parameter of the problem. Therefore has a unique fixedpoint , which is a solution of (3.5), and hence is a solution of (1.2).
Theorem 3.4.
Assume (A1)–(A4) holds. If , then (1.2) has at least one solution on .
Proof.
Choose and consider . Now define on the operators by
Let us observe that if then . Indeed it is easy to check the inequality
We can easily show that that is a contraction mapping. Let . Then
where depends only on the parameter of the problem and hence is contraction. Since is continuous, then is continuous in view of (A1). Let us now note that is uniformly bounded on . This follows from the inequality
Now let us prove that is equicontinuous.
Let and . Using the fact that is bounded on the compact set (thus , we will get
which does not depend on . So is relatively compact. By the ArzelaAscoli Theorem, is compact. We now conclude the result of the theorem based on the Krasnoselkii's theorem above.