Abstract

We give some fixed point results using an ICS mapping and involving Boyd-Wong-type contractions in partially ordered metric spaces. Our results generalize, extend, and unify several well-known comparable theorems in the literature. Also, we present some examples to support our results.

1. Introduction and Preliminaries

The Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. That is why, generalizations of the Banach principle have been heavily investigated by many authors. For instance, in 1977 Jaggi [2] proved the following theorem satisfying a contractive condition of rational type.

Theorem 1.1. Let be a complete metric space. Let be a continuous mapping such that for all distinct where with . Then, has a unique fixed point.

Existence of a fixed point for contraction-type mappings in partially ordered set has been considered by Ran and Reurings [3], and they applied their results to matrix equations. Later, Nieto and Rodríguez-López [4] studied some fixed point theorems for contractive mappings in partially ordered set and applied their main theorems to obtain a unique solution for a first-order ordinary differential equation. For more works on fixed point results in partially ordered metric spaces, we refer the reader to [529].

Note that, in the context of partially ordered metric spaces, the usual contractive condition is weakened, but at the expense that the operator is monotone.

Recently, Harjani et al. [17] proved the ordered version of Theorem 1.1 as follows.

Theorem 1.2 (see [17]). Let be a partially ordered set and suppose there exists a metric such that is a complete metric space. Let be a continuous and nondecreasing mapping such that where with . If there exists with , then has a unique fixed point.

Very recently, Luong and Thuan [21] generalized Theorem 1.2 as follows.

Theorem 1.3. Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be a nondecreasing mapping such that for all distinct with where is a lower semicontinuous function with if and only if , and Also, assume either(i)is continuous or;(ii)if is a nondecreasing sequence in such that , then .
If there exists such that , then has a fixed point.

In the sequel, we give the following definition (see e.g., [30]).

Definition 1.4. Let be a metric space. A mapping is said to be ICS if is injective (also said, one to one), continuous, and has the property: for every sequence in , if is convergent then is also convergent.

Throughout this paper, the letters and will denote the set of all nonnegative real numbers and the set of all nonnegative integer numbers, respectively.

The purpose of this paper is to generalize the above results using an ICS mapping and involving some generalized weak contractions of Boyd-Wong-type [31]. Also, some examples are presented to show that our results are effective.

2. Main Result

First, denote by the set of functions satisfying(a) for all ,(b) is upper semicontinuous from the right [i.e., for any sequence in such that , as , we have ].

Now we prove our first result.

Theorem 2.1. Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be such that is an ICS mapping and a nondecreasing mapping satisfying for all distinct with where and Also, assume either(i) is continuous or;(ii)if is a nondecreasing sequence in such that , then .
If there exists such that , then has a fixed point.

Proof. Given , define a sequence in as follows:
Since is a nondecreasing mapping, together with , we have . Inductively, we obtain
Assume that there exists such that . Since , then has a fixed point which ends the proof.
Suppose that for all . Thus, by (2.4) we have Since (2.5) holds, then condition (2.1) implies that where,
Suppose that for some . Then, inequality (2.6) turns into Regarding (2.5) and the property (a) of , we get which is a contradiction. Thus, for all . Therefore, the inequality (2.6) yields that Consequently, the sequence of positive real numbers is decreasing and bounded below. So, there exists such that .
We claim that . Suppose to the contrary that . Letting in (2.10) and using the fact that is upper semicontinuous from the right, we get which is a contradiction. Hence, we conclude that , that is,
We prove that the sequence is Cauchy in . Suppose, to the contrary, that is not a Cauchy sequence. So, there exists such that where and are subsequences of with Moreover, is chosen as the smallest integer satisfying (2.13). Thus, we have By the triangle inequality, we get Letting in above inequality and using (2.12), we get that By a triangle inequality, we have Using (2.12), (2.17) and letting in (2.18), we get Regarding , we have . From (2.1) we have where, Letting in (2.20) (and hence in (2.21)), and using (2.12), (2.17) and (2.19), we obtain which is a contradiction. Thus, is a Cauchy sequence in . Since is a complete metric space, there exists such that . Since is an ICS mapping, there exists such that But is continuous, hence
We will show that is a fixed point of .
Assume that (i) holds. Then by continuity of , we have Suppose that (ii) holds. Since is a nondecreasing sequence and then . Hence, for all . Regarding that is a nondecreasing mapping, we conclude that , or equivalently, and as , we get .
To this end, we construct a new sequence as follows: Since , so we have and hence similarly we may find that is a nondecreasing sequence. By repeating the discussion above, one can conclude that is a Cauchy sequence. Thus, converges and since is an ICS mapping, so there exists such that . The mapping is continuous, hence By (ii), we have and so we have . From (2.26), we get
If , then the proof ends.
Suppose that and since the mapping is one to one, we have , so . On account of (2.29), the expression (2.1) implies that where, Letting in (2.30) and using (2.24), (2.28), we obtain which is a contradiction. So and we have , then .

Corollary 2.2. Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be such that is an ICS mapping and is a nondecreasing mapping satisfying for all distinct with . Also, assume either(i) is continuous or;(ii)if is a nondecreasing sequence in such that , then .
If there exists such that , then has a fixed point.

Proof. Take for all in Theorem 2.1.

Corollary 2.3. Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be such that is an ICS mapping and is nondecreasing mapping with for all distinct with where with . Also, assume either(i) is continuous or;(ii)if is a nondecreasing sequence in such that , then .
If there exists such that , then has a fixed point.

Proof. Take for all in Corollary 2.2. Indeed, and this completes the proof.

Theorem 2.4. In addition to hypotheses of Theorem 2.1, assume that then has a unique fixed point.

Proof. Suppose, to the contrary, that and are fixed points of where . By (2.36), there exists a point which is comparable to and . Without loss of generality, we choose . We construct a sequence as follows: Since is nondecreasing, implies . By induction, we get .
If for some then for all . So . Analogously, we get that which completes the proof.
Consider the other case, that is, for all . Having in mind, is one to one, so for any . Then, by (2.1), we observe that where, Thus, Consequently, is a decreasing sequence of positive real numbers which is bounded below. So, there exists such that .
We claim that . Suppose, to the contrary, that . Taking and using a property of , we get which is a contradiction. Hence, we conclude that , that is, Analogously, repeating the same work we find that By uniqueness of limit of , we deduce that and since is one to one, we have , which is a contradiction. This ends the proof.

Recently, Jachymski [32] in his interesting paper showed the equivalence between several generalized contractions on (ordered) metric spaces. Since the key of his study is [32, Lemma 1], then we will combine this lemma with Theorems 2.1 and 2.4 to deduce the following.

Corollary 2.5. Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be such that is an ICS mapping and a nondecreasing mapping satisfying for all distinct with , where are continuous and nondecreasing functions such that . Also, suppose that (2.36) holds and assume either(i) is continuous or;(ii)if is a nondecreasing sequence in such that , then .
If there exists such that , then has a fixed point. Moreover, if for any , there is that is comparable to and , then has a unique fixed point.

Proof. By [32, Lemma  1], (ii) (viii), so there exists a continuous and nondecreasing function such that for all and for all distinct with . Therefore, Theorems 2.1 and 2.4 give, respectively, existence and uniqueness of the fixed point of , which completes the proof.

The following remarks are in order.(i)Corollary 2.5 corresponds to Theorems 2.1 and 2.4 of Luong and Thuan [21] by taking and for all and .(ii)Corollary 2.3 corresponds to Theorems 2.2 and 2.3 of Harjani et al. [17] by taking .

Now we give some examples illustrating our obtained results.

Example 2.6. Let be endowed with the metric for all and the order given as follows: Take the mappings be given by Set . It is easy that is nondecreasing with respect to . First, satisfies the property: if is a nondecreasing sequence in such that , then . Indeed, let be a nondecreasing sequence in with respect to such that as . We have for all .(i)If , then . From the definition of , we have . By induction, we get for all and . Then, for all and .(ii)If , then . From the definition of , we have . By induction, we get for all . Suppose that there exists such that . From the definition of , we get for all . Thus, we have and for all . Now, suppose that for all . In this case, we get and for all and .(iii)If , then . From the definition of , we have . Repeating the same idea as previous case, we get that .

Thus, we proved that in all cases, we have . We will show that (2.1) holds for all with . The unique possibilities are , and .

Case 1. If and , we have , so (2.1) holds.

Case 2. If and , we have and , so .

Case 3. If and , we have and , so .

Also, it is obvious that is an ICS mapping. All hypotheses of Theorem 2.1 are satisfied and is a fixed point of .

On the other hand, taking and , we have for each given in Theorem 1.3. Hence, the main result of Luong and Thuan [21] is not applicable.

Moreover, taking and , we have for each , , so (1.2) fails. Then, we couldn’t apply Theorem 1.1 (also, the same for Theorems 2.2 and 2.3 of Harjani et al. [17]).

Example 2.7. Let with usual order. Define by the formulas Let be given by Define by and . Then (1) is a complete ordered metric space;(2) is non-decreasing;(3) is continuous;(4) is an ICS mapping; (5) for any distinct with .

Proof. The proof of (1) and (2) is clear. To prove (3), let be a sequence in such that . By the definition of the metric , there exists such that for all . So for all . Hence . So is continuous. To prove (4), it is clear that is injective and continuous. Now, let be any sequence in such that converges to some . Then there exists such that for all . Thus for all . Hence . So is an ICS mapping. To prove (5), given with . If , then Since we have .
If and , then and . Since
we have . Then and satisfy all conditions of Corollary 2.5, so has a unique fixed point, which is .

Finally, we give a simple example which shows that if is not an ICS mapping then the conclusion of Theorem 2.1 fails.

Example 2.8. Let with the usual metric and the usual ordering. Take and for all and . The mapping is nondecreasing and continuous. Also, and there exists such that .
Let be such that for all , then is not an ICS mapping. Obviously, the condition (2.1) holds. However, has no fixed point.