Throughout the rest of this paper, we denote by a complete partially ordered metric-like space, i.e., ⪯ is a partial order on the set X and σ is a complete metric-like on X.
A mapping is said to be nondecreasing if , .
Theorem 1 Let be a complete partially ordered metric-like space. Let be a continuous and nondecreasing mapping such that for all comparable
(2.1)
where
M
is given by
and
-
(a)
is a continuous monotone nondecreasing function with if and only if ;
-
(b)
is a lower semi-continuous function with if and only if .
If there exists with , then F has a fixed point.
Proof Since F is a nondecreasing function, we obtain by induction that
Put . Then, for each integer , as the elements and are comparable, from (2.1) we get
(2.2)
which implies . Using the monotone property of the ψ-function, we get
(2.3)
Now, from the triangle inequality, for σ we have
If , then . By (2.2) it furthermore implies that
which is a contradiction. So, we have
(2.4)
Therefore, the sequence is monotone nonincreasing and bounded. Thus, there exists such that
(2.5)
We suppose that . Then, letting in the inequality (2.2), we get
which is a contradiction unless . Hence,
(2.6)
Next we prove that is a σ-Cauchy sequence. Suppose that is not a σ-Cauchy sequence. Then, there exists for which we can find subsequences and of with such that
(2.7)
Further, corresponding to , we can choose in such a way that it is the smallest integer with satisfying (2.7). Then
(2.8)
Using (2.7), (2.8) and the triangle inequality, we have
Letting and using (2.6), we obtain
(2.9)
Again, the triangle inequality gives us
Then we have
Letting in the above inequality and using (2.6) and (2.9), we get
(2.10)
Similarly, we can show that
(2.11)
As
using (2.6) and (2.9)-(2.11), we have
(2.12)
As and and are comparable, setting and in (2.1), we obtain
Letting in the above inequality and using (2.9) and (2.12), we get
which is a contradiction as . Hence is a σ-Cauchy sequence. By the completeness of X, there exists such that , that is,
(2.13)
Moreover, the continuity of F implies that
and this proves that z is a fixed point. □
Notice that the continuity of F in Theorem 1 is not necessary and can be dropped.
Theorem 2 Under the same hypotheses of Theorem 1 and without assuming the continuity of F, assume that whenever is a nondecreasing sequence in X such that implies for all , then F has a fixed point in X.
Proof Following similar arguments to those given in Theorem 1, we construct a nondecreasing sequence in X such that for some . Using the assumption of X, we have for every . Now, we show that . By (2.1), we have
(2.14)
where
Taking limit as , by (2.13), we obtain
Therefore, letting in (2.14), we get
which is a contradiction unless . Thus . □
Next theorem gives a sufficient condition for the uniqueness of the fixed point.
Theorem 3 Let all the conditions of Theorem 1 (resp. Theorem 2) be fulfilled and let the following condition be satisfied: For arbitrary two points , there exists which is comparable with both x and y. Then the fixed point of F is unique.
Proof Suppose that there exist which are fixed points. We distinguish two cases.
Case 1. If x is comparable to z, then is comparable to for and
(2.15)
where
(2.16)
Using (2.15) and (2.16), we have
which is a contradiction unless . This implies that .
Case 2. If x is not comparable to z, then there exists comparable to x and z. The monotonicity of F implies that is comparable to and , for . Moreover,
(2.17)
where
(2.18)
for n sufficiently large, because and when . Similarly as in the proof of Theorem 1, it can be shown that . It follows that the sequence is nonnegative decreasing and, consequently, there exists such that
We suppose that . Then letting in (2.17), we have
which is a contradiction. Hence . Similarly, it can be proved that
Now, passing to the limit in , it follows that , so , and the uniqueness of the fixed point is proved. □
Now, we present an example to support the useability of our results.
Example 2 Let and a partial order be defined as whenever , and define as follows:
Then is a complete partially ordered metric-like space.
Let be defined by
Define by and . We next verify that the function F satisfies the inequality (2.1). For that, given with , so . Then we have the following cases.
Case 1. If , , then
and
As , the inequality (2.1) is satisfied in this case.
Case 2. If , , then
and
As , the inequality (2.1) is satisfied in this case.
Case 3. If , , then as and , the inequality (2.1) is satisfied in this case.
Case 4. If , , then as and , the inequality (2.1) is satisfied in this case.
Case 5. If , , then as and , the inequality (2.1) is satisfied in this case.
Case 6. If , , then as and , the inequality (2.1) is satisfied in this case.
So, F, ψ and ϕ satisfy all the hypotheses of Theorem 1. Therefore F has a unique fixed point. Here 2 is the unique fixed point of F.
If we take in Theorem 1, we have the following corollary.
Corollary 1 Let be a complete partially ordered metric-like space. Let be a nondecreasing mapping such that for all comparable with
where
M
is given by
is lower semi-continuous, and if and only if . If there exists with and one of the following two conditions is satisfied:
-
(a)
F is continuous in ;
-
(b)
is a nondecreasing sequence in X such that implies for all .
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points , there exists which is comparable with both x and y, then the fixed point of F is unique.
If we take for in Corollary 1, we have the following corollary.
Corollary 2 Let be a complete partially ordered metric-like space. Let be a nondecreasing mapping such that for all comparable
where
M
is given by
and . If there exists with and one of the following two conditions is satisfied:
-
(a)
F is continuous in ;
-
(b)
is a nondecreasing sequence in X such that implies for all .
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points , there exists which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.7 in [33].
Corollary 3 Let be a complete partially ordered metric-like space. Let be a nondecreasing mapping such that for all comparable
where is a lower semi-continuous, and if and only if . If there exists with and one of the following two conditions is satisfied:
-
(a)
F is continuous in ;
-
(b)
is a nondecreasing sequence in X such that implies for all .
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points , there exists which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.1 in [10].
Corollary 4 Let be a complete partially ordered partial metric space. Let be a nondecreasing mapping such that for all comparable
where
M
is given by
, ψ is continuous monotone nondecreasing, ϕ is lower semi-continuous, and if and only if . If there exists with and one of the following two conditions is satisfied:
-
(a)
F is continuous in ;
-
(b)
is a nondecreasing sequence in X such that implies for all .
Then F has a fixed point. Moreover, the set of fixed points of F is well ordered if and only if F has one and only one fixed point.