A common fixed point theorem using implicit relation and property (E.A) in metric spaces
Summary (1 min read)
1. Introduction
- Similarly, noncompatible mapping is generalized by Aamri and Moutawakil [1] called property (E.A).
- Noncompatibility is also important to study the fixed point theory.
- There may be pairs of mappings which are noncompatible but weakly compatible (see Example 1 of Popa [6] p. 34, and Example 2.1below).
- Received: March 30, 2007 noncompatible if there exists a sequence{xn} in X such that limn→.
3. Implicit relation
- Let R and R+ denote the set of real and non-negative real numbers, respectively.
- Let us discuss property (E.A) and MKC condition for various cases.
- For every sequence {xn} in X, the authors get limn→.
- In their main Theorem the authors will apply property (E.A) and implicit relation.
- Meir-Keeler type contractive condition of Theorem A, the authors will impose property (E.A).
4. Main Results
- In this section the authors state and prove their main result.
- Assume that one of the following conditions hold: (v) {Byn} is a bounded sequence for every {yn} ⊆ X such that{Tyn} is convergent (in case (A,S) satisfies property (E.A)), and {Ayn} is a bounded sequence for every {yn} ⊆ X such that{Syn} is convergent (in case (B, T ) satisfies property (E.A)).
B, S and T have a unique common fixed point.
- Suppose that (B, T ) satisfies property (E.A) then, by definition, there exists a sequence {xn}in X such that limn→.
- Hence, in both cases, {Ayn} is a bounded sequence and {d(Ayn, Bxn)} is bounded,hencelimsupn→∞d(Ayn, Bxn) is a finite number.
- Indeed, in this case, limsupn→∞pn = limsupn→∞qn.
- The weak compatibility of A and S implies that ASu = SAu so that AAu = ASu = SAu = SSu.
- The authors can relax continuity ofF , extending this result to the more general case whereF is continuous at certain points of the boundary of (R+)6.
X, then A, B and S have a unique common fixed point.
- It is clear that two non compatible self-maps also satisfy property(E.A).
- The following remarks and examples validate their main Theorem 4.1. Remark 4.10.
- A similar reasoning provides the validity of (ĈS) and (H1).
- Thus both pairs satisfy property (E.A) and condition (iv) is satisfied for both pairs.
- The authors thank the referee’s comments on the paper.
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References
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"A common fixed point theorem using ..." refers background in this paper
...1) compatible [3] if limn→∞d(ASxn, SAxn) = 0, (1....
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...The concept of weakly commuting mappings of Sessa [7] is sharpened by Jungck [3] and further generalized by Jungck and Rhoades [4]....
[...]
...Introduction The concept of weakly commuting mappings of Sessa [7] is sharpened by Jungck [3] and further generalized by Jungck and Rhoades [4]....
[...]
653 citations
"A common fixed point theorem using ..." refers background in this paper
...Mappings A and S are said to be weakly commuting[7] if d(SAx, ASx) ≤ d(Ax, Sx), for all x ∈ X, (1....
[...]
...The concept of weakly commuting mappings of Sessa [7] is sharpened by Jungck [3] and further generalized by Jungck and Rhoades [4]....
[...]
...Introduction The concept of weakly commuting mappings of Sessa [7] is sharpened by Jungck [3] and further generalized by Jungck and Rhoades [4]....
[...]
547 citations
"A common fixed point theorem using ..." refers background or methods or result in this paper
...Therefore our main Theorem 4.1 extends Theorem B of Aamri and Moutawakil, since they considered ψ to be nondecreasing and such that 0 < ψ(t) < t,t > 0, which clearly implies ψ(0) = 0....
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...([1]) Let A, B, S and T be self-mappings of a metric space (X, d) such that (a) d(Ax,By) ≤ φ(max {d(Sx, Ty), d(By, Sx), d(By, Ty)}, ∀(x, y) ∈ X(2), where φ : R+ → R+ is a non-decreasing function on R+ such that 0 < φ(t) < t, for each t ∈ (0,∞)....
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...Compare it with condition (ii) in Theorem 1 [1]....
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...Our theorem improves and generalizes the main Theorems of Popa [6] and Aamri and Moutawakil [1]....
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...A) of [1] and implicit relation of [6] to unify under property(E....
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132 citations
"A common fixed point theorem using ..." refers background in this paper
...For a related problem, see Theorem 1 and Corollary 3 in [2]....
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64 citations
"A common fixed point theorem using ..." refers background in this paper
...F (d(Ax, By), d(x, y), d(Ax, x), d(By, y), d(By, x), d(Ax, y)) ≤ 0, for all x, y ∈ X, where F must satisfy all the conditions of implicit relation (see [8])....
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...≤ ψ(G(max{d(x, y), d(Ax, x), d(By, y), 1 2 (d(By, x) + d(Ax, y))})), for all x, y ∈ X, which is similar to condition in Theorem 1 [8]....
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...As a particular case, we can consider G(u) = ∫ u 0 φ(s) ds, where φ ≥ 0 is Lebesgue-integrable and such that ∫ 2 0 φ(t)dt > 0, for every 2 > 0, and ψ(t) = kt, with 0 ≤ k < 1 (see [8])....
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Frequently Asked Questions (14)
Q2. What is the condition that satisfies the property (E.A)?
If condition (vi) holds, and the authors suppose that {Ayn} is not bounded, then there exists a subsequence {nk} of nonnegative integer numbers such that d(Aynk , Bxnk) → +∞, as k → ∞.
Q3. what is the irrational number of x and y?
If x is an irrational number of X and y ∈ X, the authors have:F (0, |3−y|, 1, |y−2|, 1, |y−2|) = 0−hmax{2|3−y|, 1, |y−2|, 1, |y−2|} ≤ 0,as 0 ≤ h < 12 .
Q4. what is the irrational number of x?
If x is a rational number of X and y ∈ X, then the authors have: F (0, |2− y|, 0, |2− y|, 0, |2− y|) = 0− hmax{2|2− y|, 0, |2− y|, 0, |2− y|} = −2h|2− y| ≤ 0, as 0 ≤ h < 12 .
Q5. What is the meaning of a weakly compatible mapping?
Let A and S be two self-maps of a metric space (X, d) then they are said to satisfy property (E.A), if there exists a sequence {xn} in X such that limn→
Q6. What is the meaning of weakly compatible mappings?
Mappings A and S are said to be weakly commuting[7] if d(SAx, ASx) ≤ d(Ax, Sx), for all x ∈ X, (1.1) compatible [3] iflimn→∞d(ASxn, SAxn) = 0, (1.2) whenever there exists a sequence {xn} in X such that limn→
Q7. what is the case for x and y?
Let us define a function F ∈ F such that F : R+6 → R where F (t1, ..., t6) = t1 − hmax{2t2, t3, t4, t5, t6}, for each t1, ..., t6 ≥ 0, where 0 ≤ h < 12 , then the authors observe that: (i) A(X) = {2} ⊆ T (X) = X and B(X) = {2} ⊆ S(X) = {2, 3}. (ii) Let us discuss two cases for the elements x, y ∈ X and obtain the implicit relation: Case I.
Q8. what is the proof of the main theorem 4.1?
From the proof of Theorem 4.1, the authors deduce that the strict ’<’ sign can be replaced by ’≤’if the authors admit that the inequality in condition (Fu) is also strict, that is, F (u, u, 0, 0, u, u) > 0, ∀u > 0.
Q9. what is the simplest way to describe a metric space?
Let A and S be two weakly compatible self-mappings of a metric space (X, d) suchthat (i) A and S satisfy property (E.A), (ii) there exists a continuous (see Remark 4.3) function F : R+6 → R inF such thatF (d(Ax,Ay), d(Sx, Sy), d(Ax, Sx), d(Ay, Sy), d(Ay, Sx), d(Ax, Sy)) < 0,for all x, y ∈ X, where F satisfies all the conditions of implicit relation.
Q10. what is the simplest example of a metric space?
Let (X, d) be a metric space and A, B, S and T be four self-mappings on Xsatisfying:F (d(Ax,By), d(Sx, Ty), d(Ax, Sx), d(By, Ty), d(By, Sx), d(Ax, Ty)) < 0,for all x, y ∈ X, where F satisfies property (Fu).
Q11. if x = 0 and y = 1n, then d( 1?
² ≤ M(x, y) < ² + δ yields² ≤ max { d ( 0,1 n) , d(0, 0), d ( 1 n , 1 n ) , 1 2 ( d ( 1 n , 0 ) + d ( 1 n , 0 ))}= 1 n < ² + δ,showing that ² ≤ d(Ax,By) = 1n , contradicting MKC condition.
Q12. ii) If x = 0 and y = 1n+1, then?
On the other hand, for a given ² > 0, there exists a δ > 0 such that for x = xn = yn = y ∈ [ 1n+1 , 1n), the authors have that ² ≤ M(x, y) < ² + δ yields² ≤ max { 0, ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ , ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ , 1 2 [∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ + ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ ]}= ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ < ² + δ,implies d(Axn, Byn) = d ( 1 n+1 , 1 n+1 ) = 0 < ².
Q13. What is the main result of the theorem?
Assume that one of the following conditions hold: (v) {Byn} is a bounded sequence for every {yn} ⊆ X such that{Tyn} is convergent (in case (A,S) satisfies property (E.A)), and {Ayn} is a bounded sequence for every {yn} ⊆ X such that{Syn} is convergent (in case (B, T ) satisfies property (E.A)).
Q14. what is the proof of a common fixed point of A and S?
Thus AAu = Au = SAu, and Au is a common fixed point of A and S. Similarly the authors can prove that Bv is a common fixed point of B and T .