Abstract
The aim of this paper is to establish the structure of -metric spaces, as a generalization of 2-metric spaces. Some fixed point results for various contractive-type mappings in the context of ordered -metric spaces are presented. We also provide examples to illustrate the results presented herein, as well as an application to integral equations.
MSC:47H10, 54H25.
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1 Introduction
The concept of metric spaces has been generalized in many directions.
The notion of a b-metric space was studied by Czerwik in [1, 2] and many fixed point results were obtained for single and multivalued mappings by Czerwik and many other authors.
On the other hand, the notion of a 2-metric was introduced by Gähler in [3], having the area of a triangle in as the inspirative example. Similarly, several fixed point results were obtained for mappings in such spaces. Note that, unlike many other generalizations of metric spaces introduced recently, 2-metric spaces are not topologically equivalent to metric spaces and there is no easy relationship between the results obtained in 2-metric and in metric spaces.
In this paper, we introduce a new type of generalized metric spaces, which we call -metric spaces, as a generalization of both 2-metric and b-metric spaces. Then we prove some fixed point theorems under various contractive conditions in partially ordered -metric spaces. These include Geraghty-type conditions, conditions using comparison functions and almost generalized weakly contractive conditions. We illustrate these results by appropriate examples, as well as an application to integral equations.
2 Mathematical preliminaries
The notion of a b-metric space was studied by Czerwik in [1, 2].
Definition 1 [1]
Let X be a nonempty set and be a given real number. A function is a b-metric on X if, for all , the following conditions hold:
(b1) if and only if ,
(b2) ,
(b3) .
In this case, the pair is called a b-metric space.
Note that a b-metric is not always a continuous function of its variables (see, e.g., [[4], Example 2]), whereas an ordinary metric is.
On the other hand, the notion of a 2-metric was introduced by Gähler in [3].
Definition 2 [3]
Let X be a nonempty set and let be a map satisfying the following conditions:
-
1.
For every pair of distinct points , there exists a point such that .
-
2.
If at least two of three points x, y, z are the same, then .
-
3.
The symmetry: for all .
-
4.
The rectangle inequality: for all .
Then d is called a 2-metric on X and is called a 2-metric space.
Definition 3 [3]
Let be a 2-metric space, and . The set is called a 2-ball centered at a and b with radius r.
The topology generated by the collection of all 2-balls as a subbasis is called a 2-metric topology on X.
Note that a 2-metric is not always a continuous function of its variables, whereas an ordinary metric is.
Remark 1
-
1.
[5] It is straightforward from Definition 2 that every 2-metric is non-negative and every 2-metric space contains at least three distinct points.
-
2.
A 2-metric is sequentially continuous in each argument. Moreover, if a 2-metric is sequentially continuous in two arguments, then it is sequentially continuous in all three arguments; see [6].
-
3.
A convergent sequence in a 2-metric space need not be a Cauchy sequence; see [6].
-
4.
In a 2-metric space , every convergent sequence is a Cauchy sequence if d is continuous; see [6].
-
5.
There exists a 2-metric space such that every convergent sequence in it is a Cauchy sequence but d is not continuous; see [6].
For some fixed point results on 2-metric spaces, the readers may refer to [5–15].
Now, we introduce new generalized metric spaces, called -metric spaces, as a generalization of both 2-metric and b-metric spaces.
Definition 4 Let X be a nonempty set, be a real number and let be a map satisfying the following conditions:
-
1.
For every pair of distinct points , there exists a point such that .
-
2.
If at least two of three points x, y, z are the same, then .
-
3.
The symmetry: for all .
-
4.
The rectangle inequality: for all .
Then d is called a -metric on X and is called a -metric space with parameter s.
Obviously, for , -metric reduces to 2-metric.
Definition 5 Let be a sequence in a -metric space .
-
1.
is said to be -convergent to , written as , if for all , .
-
2.
is said to be a -Cauchy sequence in X if for all , .
-
3.
is said to be -complete if every -Cauchy sequence is a -convergent sequence.
The following are some easy examples of -metric spaces.
Example 1 Let and if , and otherwise , where is a real number. Evidently, from convexity of function for , then by Jensen inequality we have
So, one can obtain the result that is a -metric space with .
Example 2 Let a mapping be defined by
Then d is a 2-metric on ℝ, i.e., the following inequality holds:
for arbitrary real numbers x, y, z, t. Using convexity of the function on for , we obtain that
is a -metric on ℝ with .
Definition 6 Let and be two -metric spaces and let be a mapping. Then f is said to be -continuous at a point if for a given , there exists such that and for all imply that . The mapping f is -continuous on X if it is -continuous at all .
Proposition 1 Let and be two -metric spaces. Then a mapping is -continuous at a point if and only if it is -sequentially continuous at x; that is, whenever is -convergent to x, is -convergent to .
We will need the following simple lemma about the -convergent sequences in the proof of our main results.
Lemma 1 Let be a -metric space and suppose that and are -convergent to x and y, respectively. Then we have
for all a in X. In particular, if is constant, then
for all a in X.
Proof Using the rectangle inequality in the given -metric space, it is easy to see that
and
Taking the lower limit as in the first inequality and the upper limit as in the second inequality we obtain the desired result.
If , then
and
□
3 Main results
3.1 Results under Geraghty-type conditions
In 1973, Geraghty [16] proved a fixed point result, generalizing the Banach contraction principle. Several authors proved later various results using Geraghty-type conditions. Fixed point results of this kind in b-metric spaces were obtained by Ðukić et al. in [17].
Following [17], for a real number , let denote the class of all functions satisfying the following condition:
Theorem 1 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be an increasing mapping with respect to ⪯ such that there exists an element with . Suppose that
for all and for all comparable elements , where
If f is -continuous, 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.
Proof Starting with the given , put . Since and f is an increasing function we obtain by induction that
Step I: We will show that . Since for each , then by (3.1) we have
because
Therefore, the sequence is decreasing. Then there exists such that . Suppose that . Then, letting , from (3.2) we have
So, we have and since we deduce that which is a contradiction. Hence, , that is,
Step II: As is decreasing, if , then . Since from part 2 of Definition 4, , we have for all . Since , we have
for all . For , we have , and from (3.4) we have
It implies that
Since , from the above inequality, we have
for all . From (3.4) and (3.6), we have
for all .
Now, for all with , we have
Therefore, from (3.8) and triangular inequality
This proves that for all
Step III: Now, we prove that the sequence is a -Cauchy sequence. Using the rectangle inequality and by (3.1) we have
Letting in the above inequality and applying (3.3) and (3.7) we have
Here
Letting in the above inequality we get
Hence, from (3.10) and (3.11), we obtain
Now we claim that . If, to the contrary, , then we get
Since we deduce that
which is a contradiction. Consequently, is a -Cauchy sequence in X. Since is -complete, the sequence -converges to some , that is, .
Step IV: Now, we show that z is a fixed point of f.
Using the rectangle inequality, we get
Letting and using the continuity of f, we have . Thus, z is a fixed point of f.
Step V: Finally, suppose that the set of fixed point of f is well ordered. Assume, to the contrary, that u and v are two distinct fixed points of f. Then by (3.1), we have
because
Thus, we get , a contradiction. Hence, f has a unique fixed point. The converse is trivial. □
Note that the continuity of f in Theorem 1 can be replaced by certain property of the space itself.
Theorem 2 Under the hypotheses of Theorem 1, without the -continuity assumption on f, assume that whenever is a nondecreasing sequence in X such that , one has 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.
Proof Repeating the proof of Theorem 1, we construct an increasing sequence in X such that . Using the assumption on X we have . Now, we show that . By (3.1) and Lemma 1,
where
Therefore, we deduce that . As a is arbitrary, hence, we have .
The proof of uniqueness is the same as in Theorem 1. □
If in the above theorems we take , where , then we have the following corollary.
Corollary 1 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be an increasing mapping with respect to ⪯ such that there exists an element with . Suppose that for some r, with ,
holds for each and all comparable elements , where
If f is continuous, or, for any nondecreasing sequence in X such that one has for all , then f has a fixed point. Additionally, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Corollary 2 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be an increasing mapping with respect to ⪯ such that there exists an element with . Suppose that
for each and all comparable elements , where and .
If f is continuous, or, for any nondecreasing sequence in X such that one has 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.
Proof Since
Putting , the conditions of Corollary 1 are satisfied and f has a fixed point. □
Example 3 Let and let denote the square of the area of triangle with vertices , e.g.,
It is easy to check that d is a -metric with parameter . Introduce an order ⪯ in X by
with all other pairs of distinct points in X incomparable.
Consider the mapping given by
and the function given as
Then f is an increasing mapping with for each . If is a nondecreasing sequence in X, converging to some , then for all . Finally, in order to check the contractive condition (3.1), only the case when , , is nontrivial. But then and
All the conditions of Theorem 2 are satisfied and f has two fixed points, and . Note that the condition (stated in Theorem 1 and Theorem 2) for the uniqueness of a fixed point is here not satisfied.
3.2 Results using comparison functions
Let Ψ denote the family of all nondecreasing and continuous functions such that for all , where denotes the n th iterate of ψ. It is easy to show that, for each , the following are satisfied:
-
(a)
for all ;
-
(b)
.
Theorem 3 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be an increasing mapping with respect to ⪯ such that there exists an element with . Suppose that
where
for some and for all elements , with x, y comparable. If f is -continuous, then f has a fixed point. In addition, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Since and f is an increasing function, we obtain by induction that
By letting , we have
If there exists such that , then and so we have nothing to prove. Hence, we assume that for all .
Step I. We will prove that . Using condition (3.15), we obtain
Here
Hence,
By induction, we get
As , we conclude that
From similar arguments as in Theorem 1, since is decreasing, we can conclude that
for all .
Step II. We will prove that is a -Cauchy sequence. Suppose the contrary. Then there exist and for which we can find two subsequences and of such that is the smallest index for which
This means that
From (3.19) and using the rectangle inequality, we get
Taking the upper limit as , from (3.17) and (3.18) we get
From the definition of we have
and if , by (3.17) and (3.20) we have
Now, from (3.15) we have
Again, if by (3.21) we obtain
which is a contradiction. Consequently, is a -Cauchy sequence in X. Therefore, the sequence -converges to some , that is, for all .
Step III. Now we show that z is a fixed point of f.
Using the rectangle inequality, we get
Letting and using the continuity of f, we get
Hence, we have . Thus, z is a fixed point of f.
The uniqueness of the fixed point can be proved in the same manner as in Theorem 1. □
Theorem 4 Under the hypotheses of Theorem 3, without the -continuity assumption on f, assume that whenever is a nondecreasing sequence in X such that , one has for all . Then f has a fixed point. In addition, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Following the proof of Theorem 3, we construct an increasing sequence in X such that . Using the given assumption on X we have . Now, we show that . By (3.15) we have
where
Letting in the above relation, we get
Again, taking the upper limit as in (3.22) and using Lemma 1 and (3.23) we get
So we get , i.e., . □
Corollary 3 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be an increasing mapping with respect to ⪯ such that there exists an element with . Suppose that
where and
for all elements with x, y comparable. If f is continuous, or, whenever is a nondecreasing sequence in X such that , one has 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.
Example 4 Let be ordered by , with all other pairs of distinct points incomparable. Define by
with symmetry in all variables and with when at least two of the arguments are equal. Then it is easy to check that is a complete -metric space with .
Consider the mapping given as
and a comparison function . Then f is a nondecreasing mapping w.r.t. ⪯ and there exists such that . The only nontrivial cases for checking the contractive condition (3.15) are when and , or , (or vice versa). Then we have
resp.
Hence, all the conditions of Theorem 3 are fulfilled. The mapping f has two fixed points (A and D).
3.3 Results for almost generalized weakly contractive mappings
Berinde in [18–21] initiated the concept of almost contractions and obtained many interesting fixed point theorems. Results with similar conditions were obtained, e.g., in [22] and [23]. In this section, we define the notion of almost generalized -contractive mapping and we prove some new results. In particular, we extend Theorems 2.1, 2.2 and 2.3 of Ćirić et al. in [24] to the setting of -metric spaces.
Recall that Khan et al. introduced in [25] the concept of an altering distance function as follows.
Definition 7 [25]
A function is called an altering distance function, if the following properties hold:
-
1.
φ is continuous and nondecreasing.
-
2.
if and only if .
Let be a -metric space and let be a mapping. For , set
and
Definition 8 Let be a -metric space. We say that a mapping is an almost generalized -contractive mapping if there exist and two altering distance functions ψ and φ such that
for all .
Now, let us prove our new result.
Theorem 5 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be a continuous mapping, nondecreasing with respect to ⪯. Suppose that f satisfies condition (3.24), for all elements , with x, y comparable. If there exists such that , 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.
Proof Starting with the given , define a sequence in X such that , for all . Since and f is nondecreasing, we have , and by induction
If , for some , then and hence is a fixed point of f. So, we may assume that , for all . By (3.24), we have
where
and
From (3.25)–(3.27) and the properties of ψ and φ, we get
If
then by (3.28) we have
which gives a contradiction.
If , then
therefore (3.28) becomes
Thus, is a nonincreasing sequence of positive numbers. Hence, there exists such that
Letting in (3.29), we get
Therefore,
and hence . Thus, we have
for each .
Note that if and
Then, by (3.28) and taking , we have
which gives , a contradiction.
Next, we show that is a -Cauchy sequence in X. For this purpose, we use the following relation (see (3.9) and (3.18)):
for all (note that this can obtained as is a nonincreasing sequence of positive numbers).
Suppose the contrary, that is, is not a -Cauchy sequence. Then there exist and for which we can find two subsequences and of such that is the smallest index for which
This means that
Using (3.33) and taking the upper limit as , we get
On the other hand, we have
Using (3.30), (3.31), (3.32), and taking the upper limit as , we get
Again, using the rectangular inequality, we have
and
Taking the upper limit as in the first inequality above, and using (3.30), (3.31), and (3.34) we get
Similarly, taking the upper limit as in the second inequality above, and using (3.30), (3.31), and (3.33), we get
From (3.24), we have
where
and
Taking the upper limit as in (3.39) and (3.40) and using (3.30), (3.34), (3.36), and (3.37), we get
So, we have
and
Now, taking the upper limit as in (3.38) and using (3.35), (3.42), and (3.43) we have
which further implies that
so , a contradiction to (3.32). Thus, is a -Cauchy sequence in X.
As X is a -complete space, there exists such that as , that is,
Now, using continuity of f and the rectangle inequality, we get
Letting , we get
Therefore, we have . Thus, u is a fixed point of f.
The uniqueness of fixed point can be proved as in Theorem 1. □
Note that the continuity of f in Theorem 5 can be replaced by a property of the space.
Theorem 6 Under the hypotheses of Theorem 5, without the continuity assumption on f, assume that whenever is a nondecreasing sequence in X such that , one has , for all . Then f has a fixed point in X. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Following similar arguments to those given in the proof of Theorem 5, we construct an increasing sequence in X such that , for some . Using the assumption on X, we have , for all . Now, we show that . By (3.24), we have
where
and
Letting in (3.45) and (3.46) and using Lemma 1, we get
and
Again, taking the upper limit as in (3.44) and using Lemma 1 and (3.47) we get
Therefore, , equivalently, . Thus, from (3.47) we get and hence u is a fixed point of f. □
Corollary 4 Let be a partially ordered set and suppose that there exists a -metric d on X such that is a -complete -metric space. Let be a nondecreasing continuous mapping with respect to ⪯. Suppose that there exist and such that
for all elements with x, y comparable. If there exists such that , 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.
Proof Follows from Theorem 5 by taking and , for all . □
Corollary 5 Under the hypotheses of Corollary 4, without the continuity assumption of f, let for any nondecreasing sequence in X such that we have , for all . Then f has a fixed point in X.
4 An application to integral equations
As an application of our results, inspired by [26], we will consider the following integral equation:
Consider the set of all real continuous functions on I, ordered by the natural relation
and take arbitrary real . We will use the following assumptions.
-
(I)
, and are continuous functions;
-
(II)
for ,
-
(III)
for some and all , with x and y comparable (w.r.t. ⪯),
-
(IV)
there exists such that for all .
Let be defined by
Then is a -complete -metric space, with (similarly as in Example 2). We have the following result.
Theorem 7 Let the functions h, g, F satisfy conditions (I)-(IV) and let the space satisfy the requirement that if is a sequence in X, nondecreasing w.r.t. ⪯, and converging (in d) to some , then for all . Then the integral equation (4.1) has a solution in X.
Proof Define the mapping by
Then all the conditions of Corollary 3 are fulfilled. In particular, condition (III) implies that, for all , with x, y comparable, we have
Hence, using Corollary 3, we conclude that there exists a fixed point of f, which is obviously a solution of (4.1). □
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The authors are highly indebted to the referees of this paper who helped us to improve it in several places. The fourth author is thankful to the Ministry of Education, Science and Technological Development of Serbia.
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Mustafa, Z., Parvaneh, V., Roshan, J.R. et al. -Metric spaces and some fixed point theorems. Fixed Point Theory Appl 2014, 144 (2014). https://doi.org/10.1186/1687-1812-2014-144
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DOI: https://doi.org/10.1186/1687-1812-2014-144