#### Abstract

We define a class of almost generalized cyclic -weak contractive mappings and discuss the existence and uniqueness of fixed points for such mappings. We present some examples to illustrate our results. Moreover, we state some applications of our main results in nonlinear integral equations.

#### 1. Introduction

Fixed point theory is a crucial tool in the analysis of nonlinear problems. Banach contraction mapping principle [1] is the most known result in this direction: A self-mapping on a complete metric space has a unique fixed point if there exists such that for all . In this theorem, a self-mapping is necessarily continuous. Due to its importance to fixed point theory in nonlinear analysis, desired Banach fixed point theorem have been heavily investigated by many authors (see, e.g., [2–15]). One of the remarkable generalizations of the banach contraction mapping principle was reported by Kirk et al. [16] via cyclic contraction. A mapping is called cyclic if and , where are nonempty subsets of a metric space . Moreover, is called cyclic contraction if there exists such that for all and . Notice that although a contraction is continuous, cyclic contraction need not to be. This is one of the important gains of this theorem. In this paper, the authors also introduced the following notion.

*Definition 1.1 (see [16]). *Let be a nonempty set, be a positive integer, and be a mapping. is said to be a cyclic representation of with respect to if (i) are nonempty closed sets, (ii).

Following the paper in [16], a number of fixed point theorems on cyclic representation of with respect to a self-mapping have appeared (see, e.g., [17–26]).

The concept of almost contractions were introduced by Berinde [27, 28]. It was shown in [27] that any strict contraction, the Kannan [12] and Zamfirescu [29] mappings, as well as a large class of quasicontractions, are all almost contractions. Almost contractions and its generalizations were further considered in several works like [7, 30–37]. Recently, Ćirić et al. [7] proved some fixed point results in ordered metric spaces using almost generalized contractive condition, which is given in the following definition.

*Definition 1.2. *Let be a self-mapping on a metric space . It is said to satisfy almost generalized contractive condition if there exists and such that
for all .

In this paper, we introduce a class of almost generalized cyclic -weak contractive mappings and we investigate the existence and uniqueness of fixed points for almost generalized cyclic -weak contractive type mappings. Our main result generalizes and improves some well-known theorems in the literature (see, e.g., [16, 18, 19, 24, 26]). We state some examples to illustrate our results. Furthermore, we apply our main result to analyze the existence and uniqueness of solutions for a class of nonlinear integral equations.

#### 2. Main Result

We start this section by defining two classes of real valued functions. Let be the set of functions satisfying the following conditions: (Φ_{1}) is lower semicontinuous; (Φ_{2}). We denote by the set of all continuous functions .

*Definition 2.1. *Let be a metric space. Let be a positive integer and let be nonempty subsets of and . Let be a mapping such that (I) is a cyclic representation of with respect to ,(II) there exist , and such that
for all , (with ), where
Then is called an almost generalized cyclic -weak contractive mapping.

*Remark 2.2. *Taking in the above definition, , , , and , where is a constant, we obtain an almost generalized contractive condition. Then any almost generalized contractive mapping is an almost generalized cyclic -weak contractive mapping.

Our main result is the following.

Theorem 2.3. *Let be nonempty closed subsets of a complete metric space and . Let be an almost generalized cyclic -weak contractive mapping. Then has a unique fixed point that belongs to . *

*Proof. *Let (such a point exists since ). Define the sequence in by
We will prove that
If for some , we have , then (2.4) follows immediately. So, we can suppose that for all . From the condition (I), we observe that for all , there exists such that . Then, from the condition (II), we have
On the other hand, we have
Suppose that . Using (2.5), we obtain
which implies that . From condition (), we get that , a contradiction with our assumption for all . Thus, we have , which implies that is a decreasing sequence of positive numbers. Then there exists such that
Letting in (2.5), using (2.8), the continuity of and the lower semicontinuity of , we obtain
which implies that , that is, . Thus, we proved (2.4).

Now, we will prove that is a Cauchy sequence in . Suppose that is not a Cauchy sequence. Then there exists for which we can find two sequences of positive integers and such that for all positive integers ,
Using (2.10) and the triangular inequality, we get
Thus, we have
Letting in the above inequality and using (2.4), we obtain
On the other hand, for all , there exists such that . Then (for large enough, ) and lie in different adjacently labelled sets and for certain . Using (II), we obtain
for all , that is,
for all . Now, we have
for all . Using the triangular inequality, we get
which implies from (2.13) that
Using (2.4), we have
Again, using the triangular inequality, we get
Letting in the above inequality, using (2.20) and (2.18), we get
Similarly, we have
Letting , using (2.4) and (2.18), we obtain
Similarly, we have
Now, it follows from (2.18)–(2.24) that

Letting in (2.15), using (2.25), (2.26), the continuity of and the lower semicontinuity of , we obtain
which implies that , that is, , a contradiction with . Then we deduce that is a Cauchy sequence in the metric space .

Since is complete, there exists such that
We will prove that
From condition (I), and since , we have . Since is closed, from (2.28), we get that . Again, from the condition (I), we have . Since is closed, from (2.28), we get that . Continuing this process, we obtain (2.29).

Now, we will prove that is a fixed point of . Indeed, from (2.29), since for all , there exists such that , Applying (II) with and , we obtain
for all . On the other hand, we have
Using (2.28), we obtain that
Letting in (2.30), using (2.32), the continuity of and the lower semicontinuity of , we get
which implies that , that is, is a fixed point of .

Finally, we prove that is the unique fixed point of . Assume that is another fixed point of , that is, . From the condition (I), this implies that . Then we can apply (II) for and . We obtain
Since and are fixed points of , we can show easily that . Then we get
which implies that , that is, . Thus, we proved the uniqueness of the fixed point.

#### 3. Consequences

In this section, we derive some fixed point theorems from our main result given by Theorem 2.3.

If we take and in Theorem 2.3, then we get immediately the following fixed point theorem.

Corollary 3.1. *Let be a complete metric space and satisfies the following condition: there exist , and such that
**
for all . Then has a unique fixed point. *

An immediate consequence of Corollary 3.1 is the following fixed point theorem (see Remark 2.2).

Corollary 3.2. *Let be a complete metric space and satisfies the following condition: there exist and such that
**
for all . Then has a unique fixed point. *

*Remark 3.3. *Taking in Corollary 3.1, we obtain Theorem 2.2 in [8]. Moreover, in Corollary 3.1, it is not supposed that is nondecreasing and , as in [8].

Taking , , with , in Theorem 2.3, we derive the following result.

Corollary 3.4. *Let be nonempty closed subsets of a complete metric space and suppose that satisfies the following conditions (where ): *(i)* for ; *(ii)*there exists a constant such that
for all , for . ** Then has a unique fixed point that belongs to . *

The following fixed point theorems established in [16, 25] are immediate consequences of the above result.

Corollary 3.5. *Let be nonempty closed subsets of a complete metric space and suppose that satisfies the following conditions (where ): *(i)* for ; *(ii)*there exists a constant such that
for all , for . ** Then has a unique fixed point that belongs to . *

Corollary 3.6. *Let be nonempty closed subsets of a complete metric space and suppose that satisfies the following conditions (where ): *(i)* for ; *(ii)*there exists a constant such that
for all , for . ** Then has a unique fixed point that belongs to . *

Corollary 3.7. * for ; *(ii)*there exists a constant such that
for all , for . ** Then has a unique fixed point that belongs to . *

Corollary 3.8. * for ; *(ii)*there exist with such that
for all , for . ** Then has a unique fixed point that belongs to . *

*Remark 3.9. *Taking and , we get (i)from Corollary 3.5, the Banach contraction principle [1]; (ii)from Corollary 3.6, Kannan's fixed point theorem [12]; (iii)from Corollary 3.7, Bianchini’s fixed point theorem [2] (see also [3]); (iv)from Corollary 3.8, Hardy and Rogers fixed point theorem [9].

Now, we derive a fixed point result for cyclic mappings satisfying a contractive condition of integral type.

Denote by the set of functions satisfying the following hypotheses: (Λ_{1}) is a Lebesgue integrable mapping on each compact subset of ; (Λ_{2}) for any , we have .

We have the following result.

Corollary 3.10. *Let be nonempty closed subsets of a complete metric space and suppose satisfies the following conditions (where ): *(i)* for ; *(ii)*there exist , such that
for all , for . ** Then has a unique fixed point that belongs to . *

*Proof. *It follows immediately from Theorem 2.3 by observing that the functions and belong to .

Taking in Corollary 3.10, we obtain the following result.

Corollary 3.11. *Let be nonempty closed subsets of a complete metric space and suppose satisfies the following conditions (where ): *(i)* for ; *(ii)*there exist such that
for all , for . ** Then has a unique fixed point that belongs to . *

#### 4. Some Examples

In this section, we give some examples to illustrate our obtained results.

*Example 4.1. *Let be endowed with the standard metric for all . Consider the closed subsets and defined by and . Define the mapping by
Clearly, we have and .

Now, let . We distinguish two cases.

*Case 1. *If . In this case, we have
Define the functions by
Then we have

*Case 2. *If . In this case, we have , so inequality (4.4) is satisfied.

Similarly, if , we can show that (4.4) is satisfied.

Thus, we checked that all conditions of Theorem 2.3 are satisfied (with ). We deduce that has a unique fixed point .

*Example 4.2. *Let be endowed with the standard metric for all . Consider the closed subsets and defined by and . Define the mapping by
Clearly, we have and .

Now, let with and . We have
On the other hand, we have
Then we have
Consider the functions defined by
We have
Moreover, we can show that the above inequality holds if or .

Now, all conditions of Theorem 2.3 are satisfied (with ), we deduce that has a unique fixed point .

#### 5. An Application

In this section, we apply the result given by Theorem 2.3 to study the existence and uniqueness of solutions to a class of nonlinear integral equations.

We consider the nonlinear integral equation where is a continuous function.

Let be the set of real continuous functions on . We endow with the following standard metric: It is well known that is a complete metric space. Let , let such that We suppose that for all , we have We suppose that for all , is a decreasing function, that is, Finally, we suppose that for all , for all with and or and , where is continuous nondecreasing and belongs to . Now, define the set

We have the following result.

Theorem 5.1. *Under the assumptions (5.3)–(5.7), Problem (5.1) has one and only one solution . *

*Proof. *Define the closed subsets of , , and by
Define the mapping by
We will prove that
Let , that is,
Using condition (5.6), we obtain that
The above inequality with condition (5.4) imply that
for all . Then we have .

Similarly, let , that is,
Using condition (5.6), we obtain that
The above inequality with condition (5.5) imply that
for all . Then we have . Finally, we deduce that (5.11) holds.

Now, let , that is, for all ,
This implies from condition (5.3) that for all ,
Now, using condition (5.7), we can write that for all , we have
This implies that
Using the same technique, we can show that the above inequality holds also if we take .

Now, all the conditions of Theorem 2.3 are satisfied (with and ), we deduce that has a unique fixed point , that is, is the unique solution to (5.1).

#### Acknowledgment

M. Jleli and B. Samet are supported by the Research Center, College of Science, King Saud University.