Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 302438, 7 pages http://dx.doi.org/10.1155/2013/302438
Research Article Fixed Point Theory of Weak Contractions in Partially Ordered Metric Spaces Chi-Ming Chen,1 Jin-Chirng Lee,2 and Chao-Hung Chen2 1 2
Department of Applied Mathematics, National Hsinchu University of Education, No. 521, Nandah Road, Hsinchu City, Taiwan Department of Applied Mathematics, Chung Yuan Christian University, Taiwan
Correspondence should be addressed to Chao-Hung Chen;
[email protected] Received 12 March 2013; Accepted 20 April 2013 Academic Editor: Erdal KarapΔ±nar Copyright Β© 2013 Chi-Ming Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.
1. Introduction and Preliminaries Throughout this paper, by R+ , we denote the set of all nonnegative real numbers, while N is the set of all natural numbers. Let (π, π) be a metric space, π· a subset of π, and π : π· β π a map. We say π is contractive if there exists πΌ β [0, 1) such that, for all π₯, π¦ β π·, π (ππ₯, ππ¦) β€ πΌ β
π (π₯, π¦) .
(1)
The well-known Banachβs fixed point theorem asserts that if π· = π, π is contractive and (π, π) is complete, then π has a unique fixed point in π. In nonlinear analysis, the study of fixed points of given mappings satisfying certain contractive conditions in various abstract spaces has been investigated deeply. The Banach contraction principle [1] is one of the initial and crucial results in this direction. Also, this principle has many generalizations. For instance, Alber and Guerre-Delabriere in [2] suggested a generalization of the Banach contraction mapping principle by introducing the concept of weak contraction in Hilbert spaces. In [2], the authors also proved that the result of Eslamian and Abkar [3] is equivalent to the result of Dutta and Choudhury [4]. Later, weakly contractive mappings and mappings satisfying other weak contractive inequalities have been discussed in several works, some of which are noted in [4β16]. In 2008, Dutta and Choudhury proved the following theorem.
Theorem 1 (see [4]). Let (π, π) be a complete metric space, and let π : π β π be such that π (π (ππ₯, ππ¦)) β€ π (π (π₯, π¦)) β π (π (π₯, π¦)) , πππ πππβ π₯, π¦ β π,
(2)
where π, π : R+ β R+ are continuous and nondecreasing, and π(π‘) = π(π‘) = 0 if and only if π‘ = 0. Then π has a fixed point in π. Recently, Eslamian and Abkar [3] proved the following theorem. Theorem 2 (see [3]). Let (π, π) be a complete metric space, and let π : π β π be such that π (π (ππ₯, ππ¦)) β€ πΌ (π (π₯, π¦)) β π½ (π (π₯, π¦)) , πππ πππβ π₯, π¦ β π,
(3)
where π, πΌ, π½ : R+ β R+ are such that π is continuous and nondecreasing, πΌ is continuous, π½ is lower semicontinuous, and π (π‘) β πΌ (π‘) + π½ (π‘) > 0 βπ‘ > 0, π (π‘) = 0 ππ πππ ππππ¦ ππ π‘ = 0, Then π has a fixed point in π.
πΌ (0) = π½ (0) = 0.
(4)
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In the recent, fixed point theory has developed rapidly in partially ordered metric spaces (e.g., [17β22]). In 2012, Choudhury and Kundu [23] proved the following fixed point theorem as a generalization of Theorem 2. Theorem 3 (see [23]). Let (π, β) be a partially ordered set and suppose that there exists a metric π in π such that (π, π) is a complete metric space and let π : π β π be a nondecreasing mapping such that π (π (ππ₯, ππ¦)) β€ πΌ (π (π₯, π¦)) β π½ (π (π₯, π¦)) , πππ πππβ π₯, π¦ β π π π’πβ π‘βππ‘ π₯ β π¦,
(5)
where π, πΌ, π½ : R+ β R+ are such that π is continuous and nondecreasing, πΌ is continuous, π½ is lower semicontinuous, and π (π‘) β πΌ (π‘) + π½ (π‘) > 0 βπ‘ > 0, π (π‘) = 0 ππ πππ ππππ¦ ππ π‘ = 0,
πΌ (0) = π½ (0) = 0.
(6)
Also, if any nondecreasing sequence {π₯π } in π converges to ], then one assumes that π₯π β ]
βπ β N.
(7)
If there exists π₯0 β π with π₯0 β ππ₯0 , then π and π have a coincidence point in π. In this paper, we prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.
Ξ = {π : R+ β R+ Ξ = {π : R+ β R+
such that π is continuous} ,
such that π is lower continuous} . (9)
Let π be a nonempty set, and let (π, β) be a partially ordered set endowed with a metric π. Then, the triple (π, β, π) is called a partially ordered complete metric space. We now state the main fixed point theorem for (π, π, π, π)-contractions in partially ordered metric spaces, as follows. Theorem 5. Let (π, β, π) be a partially ordered complete metric space. Let π : π β π be monotone nondecreasing, and π (π (ππ₯, ππ¦)) β€ π (π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))) β π (max {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦)}) , (10) for all comparable π₯, π¦ β π, where π β Ξ, π β Ξ¨, π β Ξ¦, and π β Ξ, and π (π‘) β π (π‘) + π (π‘) > 0 βπ‘ > 0, π (π‘) = 0 ππ πππ ππππ¦ ππ π‘ = 0,
π (0) = π (0) = 0.
(11)
Suppose that either (a) π is continuous or
2. Fixed Point Results (I)
(b) if any nondecreasing sequence {π₯π } in π converges to ], then one assumes that
We start with the following definition. Definition 4. Let (π, β) be a partially ordered set and π : π β π. Then π is said to be monotone nondecreasing if, for π₯, π¦ β π, π₯ β π¦ σ³¨β ππ₯ β ππ¦.
And we denote the following sets of functions:
(8)
Let (π, β) be a partially ordered set. π₯, π¦ β π are said to be comparable if either π₯ β π¦ or π¦ β π₯ holds. In the section, we denote by Ξ¨ the class of functions π : +3 R β R+ satisfying the following conditions: (π1 ) π is an increasing, continuous function in each coordinate; (π2 ) for all π‘ β R+ , π(π‘, π‘, π‘) β€ π‘, π(0, 0, π‘) β€ π‘ and π(π‘, 0, 0) β€ π‘. Next, we denote by Ξ¦ the class of functions π : R+ β R+ satisfying the following conditions: (π1 ) π is a continuous, nondecreasing function; (π2 ) π(π‘) > 0 for π‘ > 0 and π(0) = 0; (π3 ) π is subadditive; that is, π(π‘1 + π‘2 ) β€ π(π‘1 ) + π(π‘2 ) for all π‘1 , π‘2 > 0.
π₯π β ] βπ β N.
(12)
If there exists π₯0 β π with π₯0 β ππ₯0 , then π has a fixed point in π. Proof. Since π is nondecreasing, by induction, we construct the sequence {π₯π } recursively as π₯π = ππ π₯0 = ππ₯πβ1
βπ β N.
(13)
Thus, we also conclude that π₯0 β π₯1 = ππ₯0 β π₯2 = ππ₯1 β β
β
β
β π₯π = ππ₯πβ1 β β
β
β
. (14) If any two consecutive terms in (14) are equal, then the π has a fixed point, and hence the proof is completed. So we may assume that π (π₯πβ1 , π₯π ) =ΜΈ 0,
βπ β N.
(15)
Now, we claim that π(π₯π , π₯π+1 ) β€ π(π₯πβ1 , π₯π ) for all π β N. If not, we assume that π(π₯πβ1 , π₯π ) < π(π₯π , π₯π+1 ) for some π β N;
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substituting π₯ = π₯π and π¦ = π₯π+1 in (10) and using the definition of the function π, we have
By letting π β β. we get that lim π (π₯ππ , π₯ππ ) = π.
π (π (π (π₯π , π₯π+1 )) , π (π (π₯π , ππ₯π )) , π (π (π₯π+1 , ππ₯π+1 )))
πββ
= π (π (π (π₯π , π₯π+1 )) , π (π (π₯π , π₯π+1 )) , π (π (π₯π+1 , π₯π+2 ))) β€ π (π (π₯π+1 , π₯π+2 )) ,
(24)
On the other hand, we have π (π₯ππ , π₯ππ )
π (max {π (π₯π , π₯π+1 ) , π (π₯π , ππ₯π ) , π (π₯π+1 , ππ₯π+1 )})
β€ π (π₯ππ , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ ) ,
= π (max {π (π₯π , π₯π+1 ) , π (π₯π , π₯π+1 ) , π (π₯π+1 , π₯π+2 )})
π (π₯ππ β1 , π₯ππ β1 )
= π (π (π₯π+1 , π₯π+2 )) , (16)
(25)
β€ π (π₯ππ β1 , π₯ππ ) + π (π₯ππ , π₯ππ ) + π (π₯ππ , π₯ππ β1 ) . Letting π β β, then we get
and hence π (π (π₯π+1 , π₯π+2 ))
lim π (π₯ππ β1 , π₯ππ β1 ) = π.
= π (π (ππ₯π , ππ₯π+1 ))
(17)
β€ π (π (π₯π+1 , π₯π+2 )) β π (π (π₯π+1 , π₯π+2 )) . Since π(π‘) β π(π‘) + π(π‘) > 0 for all π‘ > 0, we have that π(π₯π+1 , π₯π+2 ) = 0, which contradicts (15). Therefore, we conclude that π (π₯π , π₯π+1 ) β€ π (π₯πβ1 , π₯π )
βπ β N.
(18)
From the previous argument, we also have that for each π β N π (π (π₯π , π₯π+1 )) β€ π (π (π₯πβ1 , π₯π )) β π (π (π₯πβ1 , π₯π )) . (19) It follows in (18) that the sequence {π(π₯π , π₯π+1 )} is monotone decreasing; it must converge to some π β₯ 0. Taking limit as π β β in (19) and using the continuities of π and π and the lower semicontinuous of π, we get π (π) β€ π (π) β π (π) ,
(20)
which implies that π = 0. So we conclude that lim π (π₯π , π₯π+1 ) = 0.
πββ
(21)
(22)
Further, corresponding to ππ β₯ π, we can choose ππ in such a way that it the smallest integer with ππ > ππ β₯ π and π(π₯ππ , π₯ππ ) β₯ π. Therefore π(π₯ππ , π₯ππ β1 ) < π. Now we have that for all π β N π β€ π (π₯ππ , π₯ππ ) β€ π (π₯ππ , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ ) < π (π₯ππ , π₯ππ β1 ) + π.
(26)
By (14), we have that the elements π₯ππ and π₯ππ are comparable. Substituting π₯ = π₯ππ β1 and π¦ = π₯ππ β1 in (10), we have that, for all π β N, π (π (π (π₯ππ β1 , π₯ππ β1 )) , π (π (π₯ππ β1 , ππ₯ππ β1 )) , π (π (π₯ππ β1 , ππ₯ππ β1 ))) β€ π (π (π (π₯ππ β1 , π₯ππ β1 )) , π (π (π₯ππ β1 , π₯ππ )) , π (π (π₯ππ β1 , π₯ππ ))) , π (π₯ππ β1 , π₯ππ β1 )
(27)
= max {π (π₯ππ β1 , π₯ππ β1 ) , π (π₯ππ β1 , ππ₯ππ β1 ) , π (π₯ππ β1 , ππ₯ππ β1 )} = max {π (π₯ππ β1 , π₯ππ β1 ) , π (π₯ππ β1 , π₯ππ ) ,
We next claim that {π₯π } is a Cauchy sequence; that is, for every π > 0, there exists π β N such that if π, π β₯ π, then π(π₯π , π₯π ) < π. Suppose, on the contrary, that there exists π > 0 such that, for any π β N, there are ππ , ππ β N with ππ > ππ β₯ π satisfying π (π₯ππ , π₯ππ ) β₯ π.
πββ
π (π₯ππ β1 , π₯ππ )} . By the previous argument and using inequality (10), we can conclude that π (π) β€ π (π (π) , 0, 0) β π (π) β€ π (π) β π (π) ,
(28)
which implies that π = 0, a contradiction. Therefore, the sequence {π₯π } is a Cauchy sequence. Since π is complete, there exists ] β π such that lim π₯ πββ π
= ].
(29)
Suppose that (a) holds. Then (23)
] = lim π₯π+1 = lim ππ₯π = π]. πββ
πββ
Thus, ] is a fixed point in π.
(30)
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Suppose that (b) holds; that is, π₯π β ] for all π β N. Substituting π₯ = π₯π and π¦ = ] in (10), we have that
Next, we denote by Ξ the class of functions π : R+ β R+ satisfying the following conditions:
π (π (π₯π+1 , π])) = π (π (ππ₯π , π])) β€ π (π (π (π₯π , ])) , π (π (π₯π , ππ₯π )) , π (π (], π])))
(31)
Taking limit as π β β in equality (31), we have π (π (], π])) β€ π (π (0) , π (π) , π (π (], π]))) β π (π (], π])) β€ π (π (], π])) β π (π (], π])) , (32) which implies that π(], π]) = 0; that is ] = π]. So we complete the proof. If we let π (π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))) = max {π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))} ,
(π1 ) π is continuous and nondecreasing; (π2 ) for π‘ > 0, π(π‘) > 0 and π(0) = 0. And we denote by Ξ¦ the class of functions π : R+ β R+ satisfying the following conditions.
β π (max {π (π₯π , ]) , π (π₯π , ππ₯π ) , π (], π])}) .
(33)
(π1 ) π is continuous; (π2 ) for π‘ > 0, π(π‘) > 0 and π(0) = 0. We now state the main fixed point theorem for the (π, π, π)-contractions in partially ordered metric spaces, as follows. Theorem 7. Let (π, β, π) be a partially ordered complete metric space, and let π : π β π be monotone nondecreasing, and π (π (ππ₯, ππ¦)) β€ π (π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦)))
it is easy to get the following theorem. Theorem 6. Let (π, β, π) be a partially ordered complete metric space. Let π : π β π be monotone nondecreasing, and
(37)
β π (π (π₯, π¦)) + πΏ β
π (π₯, π¦) , for all comparable π₯, π¦ β π and π β Ξ¨, π β Ξ, π β Ξ¦, where πΏ > 0 and π (π₯, π¦) = max {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦)} ,
π (π (ππ₯, ππ¦)) β€ max {π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))} (34)
for all comparable π₯, π¦ β π, where π β Ξ, π β Ξ¦, and π β Ξ, and π (π‘) β π (π‘) + π (π‘) > 0 βπ‘ > 0, ππ πππ ππππ¦ ππ π‘ = 0,
π (0) = π (0) = 0.
π (π₯, π¦) = min {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦) , π (π₯, ππ¦) ,
β π (max {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦)}) ,
π (π‘) = 0
(π2 ) for π‘ β R+ , π(π‘, π‘, π‘) β€ π‘, π(π‘, 0, 0) β€ π‘, and π(0, 0, π‘) β€ π‘.
(35)
Suppose that either
π (π¦, ππ₯)} . Suppose that either (a) π is continuous or (b) if any nondecreasing sequence {π₯π } in π converges to ], then one assumes that π₯π β ] βπ β N.
(a) π is continuous or (b) if any nondecreasing sequence {π₯π } in π converges to ], then one assumes that π₯π β ]
βπ β N.
(36)
If there exists π₯0 β π with π₯0 β ππ₯0 , then π has a fixed point in π.
3. Fixed Point Results (II) In the section, we denote by Ξ¨ the class of functions π : 3 R+ β R+ satisfying the following conditions: (π1 ) π is an increasing and continuous function in each coordinate;
(38)
(39)
If there exists π₯0 β π with π₯0 β ππ₯0 , then π has a fixed point in π. Proof. If ππ₯0 = π₯0 , then the proof is finished. Suppose that π₯0 β ππ₯0 . Since π is nondecreasing, by induction, we construct the sequence {π₯π } recursively as π₯π = ππ π₯0 = ππ₯πβ1
βπ β N.
(40)
Thus, we also conclude that π₯0 β π₯1 = ππ₯0 β π₯2 = ππ₯1 β β
β
β
β π₯π = ππ₯πβ1 β β
β
β
. (41) We now claim that lim π (π₯π , π₯π+1 ) = 0.
πββ
(42)
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Put π₯ = π₯πβ1 and π¦ = π₯π in (37). Note that
Suppose, on the contrary, that there exists π > 0 such that, for any π β N, there are ππ , ππ β N with ππ > ππ β₯ π satisfying
π (π₯πβ1 , π₯π )
π (π₯ππ , π₯ππ ) β₯ π.
= min {π (π₯πβ1 , π₯π ) , π (π₯πβ1 , ππ₯πβ1 ) , π (π₯π , ππ₯π ) , π (π₯πβ1 , ππ₯π ) , π (π₯π , ππ₯πβ1 )} = min {π (π₯πβ1 , π₯π ) , π (π₯πβ1 , π₯π ) , π (π₯π , π₯π+1 ) ,
(43)
π (π₯πβ1 , π₯π+1 ) , π (π₯π , π₯π )}
(50)
Further, corresponding to ππ β₯ π, we can choose ππ in such a way that it the smallest integer with ππ > ππ β₯ π and π(π₯ππ , π₯ππ ) β₯ π. Therefore π(π₯ππ , π₯ππ β1 ) < π. By the rectangular inequality, we have
= 0.
π β€ π (π₯ππ , π₯ππ )
So, we obtain that
β€ π (π₯ππ , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ )
π (π (π₯π , π₯π+1 ))
(51)
< π (π₯ππ , π₯ππ β1 ) + π.
= π (π (ππ₯πβ1 , ππ₯π )) β€ π (π (π (π₯πβ1 , π₯π )) , π (π (π₯πβ1 , ππ₯πβ1 )) ,
Letting π β β, then we get
π (π (π₯π , ππ₯π ))) β π (π (π₯πβ1 , π₯π )) lim π (π₯ππ , π₯ππ ) = π.
β€ π (π (π (π₯πβ1 , π₯π )) , π (π (π₯πβ1 , π₯π )) , π (π (π₯π , π₯π+1 ))) β π (π (π₯πβ1 , π₯π )) , (44)
πββ
(52)
On the other hand, we have π (π₯ππ , π₯ππ )
where π (π₯πβ1 , π₯π )
β€ π (π₯ππ , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ β1 ) + π (π₯ππ β1 , π₯ππ ) ,
= max {π (π₯πβ1 , π₯π ) , π (π₯πβ1 , ππ₯πβ1 ) , π (π₯π , ππ₯π )} (45) = max {π (π₯πβ1 , π₯π ) , π (π₯πβ1 , π₯π ) , π (π₯π , π₯π+1 )} .
βπ β N.
(46)
If not, we assume that π(π₯πβ1 , π₯π ) β€ π(π₯π , π₯π+1 ); then π(π(π₯πβ1 , π₯π )) β€ π(π(π₯π , π₯π+1 )), since π is non-decreasing. Using inequality (44) and the conditions of the function π, we have that, for each π β N, π (π (π₯π , π₯π+1 )) β€ π (π (π₯π , π₯π+1 )) β π (π (π₯π , π₯π+1 )) , (47) which implies that π(π(π₯π , π₯π+1 )) = 0, and hence π(π₯π , π₯π+1 ) = 0. This contradicts our initial assumption. From the previous argument, we have that, for each π β N, π (π (π₯π , π₯π+1 )) β€ π (π (π₯πβ1 , π₯π )) β π (π (π₯πβ1 , π₯π )) , π (π₯π , π₯π+1 ) < π (π₯πβ1 , π₯π ) .
(48)
And since the sequence {π(π₯π , π₯π+1 )} is decreasing, it must converge to some π β₯ 0. Taking limit as π β β in (48) and by the continuity of π and π, we get π (π) β€ π (π) β π (π) ,
(53)
β€ π (π₯ππ β1 , π₯ππ ) + π (π₯ππ , π₯ππ ) + π (π₯ππ , π₯ππ β1 ) .
We now claim that π (π₯π , π₯π+1 ) < π (π₯πβ1 , π₯π ) ,
π (π₯ππ β1 , π₯ππ β1 )
(49)
and so we conclude that π(π) = 0 and π = 0. We next claim that {π₯π } is Cauchy; that is, for every π > 0, there exists π β N such that if π, π β₯ π, then π(π₯π , π₯π ) < π.
By letting π β β, we get that lim π (π₯ππ β1 , π₯ππ β1 ) = π.
πββ
(54)
Using inequalities (37), (52), and (54) and putting π₯ = π₯ππ β1 and π¦ = π₯ππ β1 , we have that π (π (π₯ππ , π₯ππ )) = π (π (ππ₯ππ β1 , ππ₯ππ β1 )) β€ π (π (π (π₯ππ β1 , π₯ππ β1 )) , π (π (π₯ππ β1 , ππ₯ππ β1 )) , π (π (π₯ππ β1 , ππ₯ππ β1 ))) β π (π (π₯ππ β1 , π₯ππ β1 )) + πΏ β
π (π₯ππ β1 , π₯ππ β1 ) = π (π (π (π₯ππ β1 , π₯ππ β1 )) , π (π (π₯ππ β1 , π₯ππ )) , π (π (π₯ππ β1 , π₯ππ ))) β π (π (π₯ππ β1 , π₯ππ β1 )) + πΏ β
π (π₯ππ β1 , π₯ππ β1 ) ,
(55)
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where
If we let π (π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦)))
π (π₯ππ β1 , π₯ππ β1 ) = max {π (π₯ππ β1 , π₯ππ β1 ) , π (π₯ππ β1 , π₯ππ ) , π (π₯ππ β1 , π₯ππ )} ,
= max {π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))} ,
(63)
it is easy to get the following theorem.
π (π₯ππ β1 , π₯ππ β1 ) = min {π (π₯ππ β1 , π₯ππ β1 ) , π (π₯ππ β1 , π₯ππ ) , π (π₯ππ β1 , π₯ππ ) , π (π₯ππ β1 , π₯ππ ) , π (π₯ππ β1 , π₯ππ )} . (56)
β π (π (π₯, π¦)) + πΏ β
π (π₯, π¦) ,
lim π (π₯ππ β1 , π₯ππ β1 ) = π,
πββ
lim π (π₯ππ β1 , π₯ππ β1 ) = 0,
πββ
(57)
This implies that π(π) = 0, and hence π = 0. So we get a contraction. Therefore {π₯π } is a Cauchy sequence. Since π is complete, there exists ] β π such that lim π₯π = ].
(58)
πββ
for all comparable π₯, π¦ β π and π β Ξ, π β Ξ¦, where πΏ > 0 and π (π₯, π¦) = max {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦)} ,
π (π) β€ π (π (π) , 0, 0) β π (π) β€ π (π) β π (π) .
Suppose that (a) holds. Then
π (π₯, π¦) = min {π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦) , π (π₯, ππ¦) , π (π¦, ππ₯)} . (65) Suppose that either (a) π is continuous or
] = lim π₯π+1 = lim ππ₯π = π]. πββ
(59)
Thus, ] is a fixed point in π Suppose that (b) holds; that is, π₯π β ] for all π β N. Substituting π₯ = π₯π and π¦ = ] in (37), we have that π (π (π₯π+1 , π])) = π (π (ππ₯π , π])) β€ π (π (π (π₯π , ])) , π (π (π₯π , ππ₯π )) , π (π (], π])))
(60)
where π (π₯π , ]) = max {π (π₯π , ]) , π (π₯π , ππ₯π ) , π (], π])} , π (π₯π , ]) = min {π (π₯π , ]) , π (π₯π , ππ₯π ) , π (], π]) , π (π₯π , π])
(b) if any nondecreasing sequence {π₯π } in π converges to ], then one assumes that π₯π β ] βπ β N.
(66)
If there exists π₯0 β π with π₯0 β ππ₯0 , then π has a fixed point in π.
Acknowledgment The authors would like to thank the referee(s) for the many useful comments and suggestions for the improvement of the paper.
β π (π (π₯π , ])) + πΏ β
π (π₯π , ]) ,
References (61)
π (], ππ₯π )} . Letting π β β, then we obtain that π (π₯π , ]) σ³¨β π (], π]) ,
π (π (ππ₯, ππ¦)) β€ max {π (π (π₯, π¦)) , π (π (π₯, ππ₯)) , π (π (π¦, ππ¦))} (64)
Letting π β β, then we obtain that
πββ
Theorem 8. Let (π, β, π) be a partially ordered complete metric space, and let π : π β π be monotone nondecreasing, and
π (π₯π , ]) σ³¨β 0,
π (π (], π])) β€ π (π (0) , π (0) , π (π (], π]))) β π (π (], π])) β€ π (π (], π])) β π (π (], π])) , (62) which implies that π(], π]) = 0; that is, ] = π]. So we complete the proof.
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