Abstract: The purpose of this paper was to prove some fixed point theorems in Sb-metric spaces by using (α, β)-admissible Geraghty type rational contractive conditions and some suitable examples have been provided with relevant to the results. Also, an application to Homotopy theory as well as integral equations were given.
INTRODUCTION
In Banch1 introduced the notion of Banach contraction principle. It is most fundamental tool in nonlinear analysis and some results related with generalization of various type of metric spaces2-5.
In recent time, Sedghi et al.6 described Sb-metric spaces by applying the concept of S and b-metric spaces and established some fixed point results in Sb-metric spaces. Subsequently to improve many author’s established numerous results on -metric spaces7-10.
In Geraghty11 studied a generalization of Banach contraction principle. In Samet et al.12 initiated the concept of α-contractive and α-admissible mappings and proved fixed point theorems on complete metric spaces for such class of mappings. In Cho et al.13 initiated the concept of α-Geraghty contractive type mappings. On the other hand, Karapinar14 established the existence of a unique fixed point for a triangular α-admissible mapping which is a generalized α-ψ-Geraghty contractive type mapping. Later on, Chandok15 illustrated the theory of (α, β)-admissible Geraghty type contractive mappings. Also very recently, Gupta et al.16 proved some fixed point results in ordered metric spaces under the (ψ, β)-admissible Geraghty contractive type mappings.
Subsequently, this type of research has been studied by several investigators17-25.
The aim of present article was to prove unique fixed point theorems for (α, β)-admissible Geraghty type contraction in ordered Sb-metric spaces. The obtained results generalized, unified and modified some recent theorems in the literature. Some suitable example and an applications to Homotopy theory as well as integral equations were given here to illustrate the usability of the obtained results.
Firstly, recall some definitions, lemmas and examples.
PRELIMINARIES
Definition: ([6]) Let: Sb: X3→[0, 1) be a mapping defined on a non-empty set X and b>1 be a given real number. Suppose that the mapping Sb satisfies the following properties:
(Sb3) 0<Sb(l, m, n)<b (Sb(l, l, x)+Sb(m, m, x)+Sb(n, n, x)) for all l, m, n∈ X. Then, the function Sb is called a Sb-metric on X and the pair (X, Sb) is called a Sb-metric space.
Remark: ([6]) It must be noted that, the class of Sb-metric spaces is definitely larger than that of S-metric spaces. In fact, each S-metric space is a Sb-metric space whenever b = 1.
Following example shows that a Sb-metric space on X need not be a S-metric spaces.
Example: ([6]) Let (X, S) be S-metric space and S*(l, m, n) = S(l, m,n)k where k>1 is a real number. Note that (X, S*) is not necessarily S-metric space but S* is a Sb-metric with b = 22(k-1).
Definition: ([6]) Let (X, Sb) be a Sb-metric space. Then, we define the open ball Bsb (l, r) and closed ball Bsb [l, r] with centre l∈ X and radius r>0 as following respectively:
and:
Lemma: ([6]) In a Sb-metric space, we have Sb(u, u, w)<2b Sb (u, u, v)+b2Sb(v, v, w).
Definition: ([6]) If (X, Sb) be a Sb-metric space. A sequence {xn} in X is said to be:
| • | Sb-Cauchy sequence if, for each ∈>0, there is an integer n0 ∈ Z+ such that Sb (xn, xn, xm)<∈ for each n, m>n0 |
| • | Sb-convergent to a point x∈ X if, for each ∈>0, there is an integer n0 ∈ Z+ such that Sb (xn, xn, x)<∈ or Sb (x, x, xn)<∈ for all n> n0 and denoted by limn→∞ xn = x |
Definition: ([6]) A Sb-metric space (X, Sb) is called complete if every Sb-Cauchy sequence is Sb-convergent in X.
Lemma: ([6]) If (X, Sb) be a Sb-metric space with b>1 and suppose that {xn} is a Sb-convergent to x, then:
and:
for all y∈ X. Specifically, if x = y then limn→∞ Sb (xn, xn, y) = 0.
Definition: Let E: X→X be a self-mapping and α, β: X×X×X→R+ defined on non-empty set X. Then, the mapping E is said to be (α, β)-admissible mapping, if α(x, x, y)>1 and β(x, x, y)>1 implies α (Ex, Ex, Ey)>1 and β (Ex, Ex, Ey)>1 for all x, y∈X.
Definition: Let (X, Sb) be a Sb-metric space, α, β: X×X×X→[0, ∞] be a mappings defined on non-empty set X. We say that X is a (α, β)-regular if {xn} is a sequence in X such that xn→x∈X, α (xn, xn, xn+1)>1 and β (xn, xn, xn+1)>1 and
and:
and
RESULTS AND DISCUSSIONS
Let Ω = {Ω/Ω: [0, ∞) →[0, 1)} be a family of function then {tn} be a any bounded sequence of positive reals such that (tn)→1 astn→0. Let Φ = {Φ: Φ: [0, ∞)→[0, ∞)} be a family of functions such that Φ is continuous, strictly increasing and Φ(0) = 0.
Definition: Let (X, Sb) be a Sb-metric space, α, β: X×X×X→R+ and E: X→X is said to be (α, β)-Geraghty type-I and type-II rational contractive mapping if, there exists Ω∈Ω such that for all x, y∈X, satisfies the following conditions:
| (1) |
| (2) |
For all x, y∈X, x is comparable to y, I = 3 or 4 and φ∈Φ, r>1. Where:
Theorem: Let (X, Sb,
| • | E is an (α, β)-admissible mapping |
| • | E is an (α, β)-Geraghty type-I rational contractive mapping with i = 4 |
| • | There exist x0 ∈X such that x0 |
| • | Either E is continuous or X is (α, β)-regular |
Then E has a unique fixed point in X.
Proof Let X0 ∈ X such that α(x0, x0, Ex0)>1 and β (x0, x0, Ex0)>1, since E is self-map, then ∃ a sequence {xn} in X such that xn+1 = E xn, n = 0, 1, 2, 3.
Case I: If xn = Exn = xn+1, then clearly xn is a fixed point of E.
Case II: Assume xn ≠ E xn, ∀n.
Since x0
Since E is (α, β)-admissible mapping, α(x0, x0, Ex0) = α(x0, x0, x1)>1:
Hence by induction, we get α(xn, xn, xn+1)>1 for all n>0.
Similarly, β (xn, xn, xn+1)>1 for all n>0.
Now:
Where:
Thus:
Also:
Continuing this way we can conclude that:
If Sb (Enx0, Enx0, En+1x0)<Sb(En+1x0, En+1x0, En+2x0), which is contradiction.
Hence Sb(En+1x0, En+1x0, En+2x0)<Sb (Enx0, Enx0, En+1x0). Thus, {Sb (Enx0, Enx0, En+1x0)} is non-increasing and must converges to a real number ξ>0. Such that limn→∞ Sb (Enx0, Enx0, En+1x0) = ξ. If ξ>0 which is contradiction. Hence ξ = 0. Thus limn→∞ Sb (Enx0, Enx0, En+1x0) = 0.
Now we prove that {Enx0} is a Cauchy sequence in (X, Sb). On contrary assume that {Enx0} is not Cauchy sequence. Then there exist ∈>0 and monotonically increasing sequence of natural numbers {mk} and {nk} such that nk>mk:
| (3) |
and:
| (4) |
Letting k→∞:
| (5) |
Where:
But:
Now from Eq. 5:
It is clear that:
Hence:
Which is contradiction. Hence {Enxo} is a Cauchy sequence in (X, Sb). Because of completeness of (X, Sb), there is an ν∈X with {Enxo} → ν∈(X, Sb).
Assume that E is continuous. Therefore:
Now, assume that X is (α- β)regular. Therefore, there exists a sub sequence {xnk} of {xn} such that:
for all k∈N and α (ν, ν, E ν)>1 and β (ν, ν, E ν)>1. Since and (X, Sb) is regular, it follows is comparable to ν.
Suppose E ν≠ν. From (1) and by the definition of φ, by known Lemma:
| (6) |
Here:
Hence from Eq. 6:
So, we have:
That is:
Consequently:
And hence Sb (ν, ν, Eν) = 0 that is ν = Eν. Therefore, ν is fixed point of E.
Further to prove the uniqueness, suppose that ν* is also anther fixed point of E such that ν ≠ ν* and α (ν, ν, E ν)>1, α (ν*, ν*, E ν*)>1 and β (ν, ν, E ν)>1, β (ν*, ν*, E ν*)>1.
Consider:
Where:
Which is contradiction. Unless Sb (ν, ν, ν*) = 0, that is ν = ν*. Hence E has a unique fixed point.
Corollary: In the hypothesis of above Theorem, replace i = 3 in place of i = 4. Then, E has a unique fixed point.
Example: Let Sb: X×X×X→R+ be a mapping defined as:
where, X = [0, ∞) and
Also define α, β: X×X×X→R+ and Ω: [0, ∞]→[0, 1), φ: [0, ∞)→[0, ∞) as:
and:
Since (X, Sb,
Hence, the given inequality is satisfied. Otherwise α(p, p, Ep) β(q, q, Eq) = 0. Then:
Therefore, all the conditions are satisfied of above Theorem and 0 is unique fixed point of E.
Theorem: Let (X, Sb,
(II) E is continuous or if an increasing sequence {xn}→x∈ X, then xn
Proof: Similar proof follows from above Theorem.
Example: Let Sb: X×X×X → R+ be a mapping defined as:
where, X= [0, ∞) and
for all p∈ X, also define Ω: [0, ∞)→[0, 1), by:
Then, by above Theorem, 0 is unique fixed point of E.
Theorem: In the hypotheses of above Theorem, replace (2) in place of (1). Then, E has a unique fixed point.
Corollary: In the hypotheses of above Theorem, replace i = 3 in place of i = 4. Then, E has a unique fixed point.
APPLICATIONS
Application to homotopy
Theorem: Let (X, Sb) be complete Sb-metric space, U and be an open and closed subset of X such that
| • | u≠,Hb (u, κ) for each u∈ ∂U and κ ∈ [0,1] (Here ∂U is boundary of U in X) |
| • | α (u, u, Hb (u, κ)) β(v, v, Hb (v, κ))φ(4b5 Sb(Hb(u, κ), Hb(u, κ), Hb(u, κ)) |
| • | <Ω (φ (Sb(u, u, v)) φ(Sb(u ,u, v)) |
For all u, v∈
| • | There exist Mb>0 such that Sb (Hb (u, κ), Hb (u, κ), Hb (u, ζ))<Mb|κ-ζ|. For every u ∈ and κ, ζ ∈ [0, 1]. Then, Hb (.,0) has a fixed point ⇔ Hb (.,1) has a fixed point |
Proof: Let the set B = {κ ∈ [0, 1]: u = Hb (u, κ) for some u ∈ U}. Since Hb (.,0) has a fixed point in U, so 0 ∈ B.
Now, prove B is closed as well as open in [0, 1] and hence by the connectedness B = [0, 1]. Let
Now, κ ∈ B must be shown. Since κn ∈ B for n = 0, 1, 2, 3,….. there exists un ∈ U with un+1 = Hb (un, κn). Since Hb is (α, β)-admissible operator:
and:
Hence by induction α(un, un, un+1)>1 for all n>0. Similarly, β(un, un, un+1)>1 for all n>0. Consider:
Letting n→∞:
By the hypothesis, we have:
Therefore:
In the above Inequality:
Since Ω∈Ω, it follows:
Which yields:
| (7) |
Now, {un} is a Sb-Cauchy sequence in (X, Sb) is to be shown. On contrary assume that {un} is not a Sb-Cauchy sequence.
There is an ε>0 and monotone increasing sequence of natural numbers {mk} and {nk} such that nk>mk:
| (8) |
and:
| (9) |
Letting k→∞:
| (10) |
But:
From Eq. 10:
That is:
Which implies:
Consequently, we obtain:
and hence:
It is a contradiction.
Hence {un} is a Sb-Cauchy sequence in (X, Sb). By completeness there exists η∈U such that:
| (11) |
Now:
So:
That is:
implies:
Consequently, we get:
and hence Sb (Hb (η, κ), Hb (η, κ), η) = 0. It follows that Hb (η, κ) = η.
Thus κ∈ B. Clearly B is closed in [0, 1]. Let κ0∈ B. Then there exists u0∈ U such that u0 = Hb (u0, κ0). Since U is open, then there exists r>0 such that Choose κ∈ (κ0-∈, κ0+∈) such that:
Then, for:
Letting n→∞ and applying φ on both sides, then:
Therefore:
Thus for each fixed κ∈ (κ0-∈, κ0+∈), Hb (, ;κ):
Similar process can be used to prove the converse.
Applications to integral equations
Theorem: Consider the I.V.P:
| (12) |
where, K:I×R→R is a continuous function and x0∈ R. Let Ω: [0, ∞)→ [0, 1), φ: [0, ∞)→[0, ∞) be a two functions defined as:
And consider the following conditions:
| • | If there exist a function θ: R3→R such that there is an x1 ∈ C(I), for all t∈I, we have: |
| • | For all t∈I and for all x, y ∈ C(I ), θ(x(t), x (t), y(t))>0⇒: |
| • | For any point x of a sequence {xn} of points in C(I) with: |
θ (xn, xn, xn+1)>0, limn→∞ inf θ (xn, xn, x)>0
Then, (12) has a unique solution.
Proof: The integral equation of I.V.P (12) is:
Let X = C (I) be the space of all continuous functions defined on I and let Sb (x, y, z) = (|y+z-2x|+|y-z|)2 for x, y, z ∈X. Then (X, Sb) is a complete Sb-metric space, also define E: X→X by:
| (13) |
Now:
Thus:
With θ(x(t), y(t))>0 for all t∈I. Define α, β: X×X×X→[0, ∞) by:
Then, obviously E is an (α, β)-admissible, for all, y∈X, then:
It follows from Eq. 2, E has a unique fixed point in X.
CONCLUSION
This study presents some fixed point results by using (α, β)-admissible Geraghty type rational contractive conditions defined on ordered Sb-metric spaces and suitable examples that supports the main results. Also, applications to Homotopy theory as well as integral equations are provided.
SIGNIFICANCE STATEMENT
This study proposed a framework to established fixed point results by using (α, β)-admissible Geraghty type rational contractions in ordered Sb-metric spaces. This study will help researchers to generalized different contractions in Sb-metric spaces with applications to integral equations as well as Homotopy theory. Thus, a new framework on Sb-metric spaces may be arrived at.
ACKNOWLEDGMENT
The authors are very thankful to the reviewers and editors for their valuable comments, remarks and suggestions which improved the paper in good form.