Volume 2011, Article ID 132367,20pages doi:10.1155/2011/132367
Research Article
Fixed Point Theorems for Generalized Weakly
Contractive Condition in Ordered Metric Spaces
Hemant Kumar Nashine
1and Ishak Altun
21Department of Mathematics, Disha Institute of Management and Technology, Satya Vihar,
Vidhansabha-Chandrakhuri Marg, Mandir Hasaud, Raipur 492101 (Chhattisgarh), India 2Department of Mathematics, Faculty of Science and Arts, Kirikkale University, 71450 Yahsihan,
Kirikkale, Turkey
Correspondence should be addressed to Ishak Altun,[email protected]
Received 30 September 2010; Accepted 11 February 2011
Academic Editor: Yeol J. Cho
Copyrightq2011 H. K. Nashine and I. Altun. 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.
Fixed point results with the concept of generalized weakly contractive conditions in complete ordered metric spaces are derived. These results generalize the existing fixed point results in the literature.
1. Introduction and Preliminaries
There are a lot of generalizations of the Banach contraction mapping principle in the literature. One of the most interesting of them is the result of Khan et al.1. They addressed a new category of fixed point problems for a single self-map with the help of a control function which they called an altering distance function.
A functionϕ:0,∞ → 0,∞is called an altering distance function ifϕis continuous, nondecreasing, andϕ0 0 holds.
Khan et al.1given the following result.
Theorem 1.1. LetX, dbe a complete metric space, letϕbe an altering distance function, and let
T:X → Xbe a self-mapping which satisfies the following inequality:
ϕdTx,Ty≤cϕdx, y, 1.1
In fact, Khan et al.1proved a more general theorem of which the above result is a corollary. Another generalization of the contraction principle was suggested by Alber and Guerre-Delabriere 2in Hilbert Spaces by introducing the concept of weakly contractive mappings.
A self-mappingTon a metric spaceXis called weakly contractive if for eachx, y∈ X,
dTx,Ty≤dx, y−φdx, y, 1.2
whereφ:0,∞ → 0,∞is positive on0,∞andφ0 0.
Rhoades 3 showed that most results of 2 are still valid for any Banach space. Also, Rhoades3proved the following very interesting fixed point theorem which contains contractions as special caseφt 1−kt.
Theorem 1.2. LetX, dbe a complete metric space. IfT:X → Xis a weakly contractive mapping, and in addition,φis continuous and nondecreasing function, thenThas a unique fixed point.
In fact, Alber and Guerre-Delabriere2assumed an additional condition onφwhich is limt→ ∞φt ∞. But Rhoades3obtained the result noted inTheorem 1.2without using this particular assumption. Also, the weak contractions are closely related to maps of Boyd and Wong4and Reich type5. Namely, ifφis a lower semicontinuous function from the right, thenψt t−φtis an upper semicontinuous function from the right, and moreover,
1.2turns intodTx,Ty≤ψdx, y. Therefore, the weak contraction is of Boyd and Wong type. And if we define βt 1−φt/t fort > 0 andβ0 0, then1.2is replaced by
dTx,Ty≤βdx, ydx, y. Therefore, the weak contraction becomes a Reich-type one. Recently, the following generalized result was given by Dutta and Choudhury 6
combiningTheorem 1.1andTheorem 1.2.
Theorem 1.3. LetX, dbe a complete metric space, and letT:X → Xbe a self-mapping satisfying the inequality
ϕdTx,Ty≤ϕdx, y−φdx, y, 1.3
for allx, y ∈ X, whereϕ,φ : 0,∞ → 0,∞are both continuous and nondecreasing functions withϕt 0φtif and only ift0. Then,Thas a unique fixed point.
Also, Zhang and Song7given the following generalized version ofTheorem 1.2.
Theorem 1.4. LetX, dbe a complete metric space, and letT,S :X → Xbe two mappings such that for eachx, y∈ X,
dTx,Sy≤Φx, y−φΦx, y, 1.4
whereφ:0,∞ → 0,∞is a lower semicontinuous function withφt>0fort >0andφ0 0,
Φx, ymax
dx, y, dx,Tx, dy,Sy,1
2
dy,Txdx,Sy. 1.5
Very recently, Abbas and Doric 8 and Abbas and Ali Khan 9 have obtained common fixed points of four and two mappings, respectively, which satisfy generalized weak contractive condition.
In recent years, many results appeared related to fixed point theorem in complete metric spaces endowed with a partial orderingin the literature10–25. Most of them are a hybrid of two fundamental principle: Banach contraction theorem and the weakly contractive condition. Indeed, they deal with a monotone either order-preserving or order-reversing mapping satisfying, with some restriction, a classical contractive condition, and such that for somex0 ∈ X, eitherx0 Tx0orTx0 x0, whereTis a self-map on metric space. The first result in this direction was given by Ran and Reurings22, Theorem 2.1who presented its applications to matrix equation. Subsequently, Nieto and Rod´riguez-L ´opez18extended the result of Ran and Reurings22for nondecreasing mappings and applied to obtain a unique solution for a first-order ordinary differential equation with periodic boundary conditions.
Further, Harjani and Sadarangani 26 proved the ordered version of Theorem 1.2, Amini-Harandi and Emami12proved the ordered version of Rich type fixed point theorem, and Harjani and Sadarangani27proved ordered version ofTheorem 1.3.
The aim of this paper is to give a generalized ordered version ofTheorem 1.4. We will do this using the concept of weakly increasing mapping mentioned by Altun and Simsek11
also see28,29.
2. Main Results
We will begin with a single map. The following theorem is a generalized version of Theorems 2.1 and 2.2 of Harjani and Sadarangani27.
Theorem 2.1. LetX,be a partially ordered set, and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetT:X → Xbe a nondecreasing mapping such that
ϕdTx,Ty≤ϕΘx, y−φΘx, y foryx, 2.1
where
Θx, yadx, ybdx,Tx cdy,Tyedy,Txdx,Ty, 2.2
a >0, b, c, e ≥0, abc2e ≤ 1,ϕ, φ:0,∞ → 0,∞, ϕis continuous, nondecreasing,φ
is lower semicontinuous functions, andϕt 0φtif and only ift0. Also, suppose that there existsx0 ∈ Xwithx0 Tx0. If
Tis continuous, 2.3
or
{xn} ⊂ X is a nondecreasing sequence withxn−→zinX, thenxnz ∀n 2.4
Proof. IfTx0 x0, then the proof is completed. Suppose thatTx0/x0. Now, sincex0 Tx0, andTis nondecreasing, we have
x0 Tx0 T2x0 · · · Tnx0 Tn1x0· · ·. 2.5
Putxn Tnx0, and soxn1 Txn. If there existsn0 ∈ {1,2, . . .}such thatΘxn0, xn0−1 0,
then it is clear thatxn0−1xn0 Txn0−1, and so we are finished. Now, we can suppose that
Θxn, xn−1>0, 2.6
for alln≥1.
First, we will prove that limn→ ∞dxn1, xn 0. From2.2, we have forn≥1
Θxn, xn−1 adxn, xn−1 bdxn,Txn cdxn−1,Txn−1
edxn−1,Txn dxn,Txn−1
acdxn, xn−1 bdxn, xn1 edxn−1, xn1
≤acedxn, xn−1 bedxn, xn1.
2.7
Now, we claim that
dxn1, xn≤dxn, xn−1, 2.8
for alln≥1. Suppose that this is not true; that is, there existsn0 ≥1 such thatdxn01, xn0>
dxn0, xn0−1. Now, sincexn0 xn01, we can use the2.1for these elements, then we have
ϕdxn01, xn0 ϕdTxn0,Txn0−1
≤ϕΘxn0, xn0−1−φΘxn0, xn0−1
≤ϕacedxn0, xn0−1 bedxn0, xn01
−φΘxn0, xn0−1
≤ϕabc2edxn0, xn01−φΘxn0, xn0−1
≤ϕdxn0, xn01−φΘxn0, xn0−1.
2.9
This impliesφΘxn0, xn0−1 0, by the property ofφ, we haveΘxn0, xn0−1 0, which this
and bounded below. Thus there existsρ ≥0 such that limn→ ∞dxn1, xn ρ. Now suppose thatρ >0. Therefore from2.2
lim
n→ ∞adxn, xn−1≤lim supn→ ∞ Θxn, xn−1
lim sup n→ ∞
acdxn, xn−1 bdxn, xn1 edxn−1, xn1
≤lim sup
n→ ∞ acedxn, xn−1 bedxn, xn1.
2.10
This implies
0< aρ≤lim sup
n→ ∞ Θxn, xn−1≤abc2eρ≤ρ 2.11
and so there existρ1>0 and a subsequence{xnk}of{xn}such that limk→ ∞Θxnk, xnk−1
ρ1≤ρ.
By the lower semicontinuity ofφwe have
φρ1
≤lim inf k→ ∞ φ
Θxnk, xnk1
. 2.12
From2.1, we have
ϕdxnk1, xnkϕdTxnk,Txnk−1
≤ϕΘxnk, xnk−1
−φΘxnk, xnk−1
, 2.13
and taking upper limit ask → ∞, we have
ϕρ≤ϕρ1
−lim inf k→ ∞ φ
Θxnk, xnk1
≤ϕρ1
−φρ1
≤ϕρ−φρ1
,
2.14
that is, φρ1 0. Thus, by the property of φ, we have ρ1 0, which is a contradiction. Therefore, we haveρ0.
Next, we show that{xn}is Cauchy.
Suppose that this is not true. Then, there is anε > 0 such that for an integerk, there exist integersmk> nk> ksuch that
dxnk, xmk> ε. 2.15
For every integerk, letmkbe the least positive integer exceedingnksatisfying2.15and such that
dxnk, xmk−1
Now,
ε < dxnk, xmk
≤dxnk, xmk−1
dxmk−1, xmk.
2.17
Then, by2.15and2.16, it follows that
lim k→ ∞d
xnk, xmk
ε. 2.18
Also, by the triangle inequality, we have
d
xnk, xmk−1
−dxnk, xmk< dxmk−1, xmk. 2.19
By using2.18, we get
lim k→ ∞d
xnk, xmk−1
ε. 2.20
Now, by2.2, we get
adxnk, xmk−1
≤Θxnk, xmk−1
adxnk, xmk−1
bdxnk,Txnkcdxmk−1,Txmk−1
,
edxmk−1,Txnk
dxnk,Txmk−1
adxnk, xmk−1
bdxnk, xnk1
cdxmk−1, xmk,
edxmk−1, xnk1
dxnk, xmk
≤adxnk, xmk−1
bdxnk, xnk1
cdxmk−1, xmk
,
edxmk−1, xnkdxnk, xnk1
dxnk, xmk,
2.21
and taking upper limit ask → ∞and using2.18and2.20, we have
0< aε≤lim sup k→ ∞ Θ
xnk, xmk−1
≤a2eε≤ε. 2.22
This implies that there existε1>0 and a subsequence{xnkp}of{xnk}such that
lim p→ ∞Θ
xnkp, xmkp−1
ε1≤ε. 2.23
By the lower semicontinuity ofφ, we have
φε≤lim inf k→ ∞ φ
Θxnk, xmk−1
Now, by2.1, we get
ϕε lim sup p→ ∞ ϕ
dxnkp, xmkp
≤lim sup p→ ∞ ϕ
dxnkp, xnkp1
dTxnkp,Txmkp−1
lim sup p→ ∞ ϕ
dTxnkp,Txmkp−1
≤lim sup p→ ∞
ϕΘxnkp, xmkp−1
−φΘxnkp, xmkp−1
ϕε1−lim infp→ ∞ φ
Θxnkp, xmkp−1
≤ϕε1−φε1
≤ϕε−φε1,
2.25
which is a contradiction. Thus,{xn}is a Cauchy sequence. From the completeness ofX, there existsz ∈ Xsuch thatxn → zasn → ∞. IfTis continuous, then it is clear thatTz z. If
2.4holds, then we havexn zfor alln. Therefore, for alln, we can use2.1forxnandz. Since
Θz, xn adz, xn bdz,Tz cdxn,Txn edxn,Tz dz,Txn
adz, xn bdz,Tz cdxn, xn1 edxn,Tz dz, xn1,
2.26
and so limn→ ∞Θz, xn bedz,Tz, we have
ϕdTz, z lim sup
n→ ∞ ϕdTz, xn1
lim sup
n→ ∞ ϕdTz,Txn
≤lim sup n→ ∞
ϕΘz, xn−φΘz, xn
≤ϕbedTz, z−φbedTz, z
≤ϕdTz, z−φbedTz, z.
2.27
By the property ofφ, we haveTzz. Thus, the proof is complete.
The following corollary is a generalized version of Theorems1.2and 1.3of Harjani and Sadarangani26.
Corollary 2.2. LetX,be a partially ordered set, and suppose that there exists a metricdin X such thatX, dis a complete metric space. LetT:X → Xbe a nondecreasing mapping such that
where
Θx, yadx, ybdx,Tx cdy,Tyedy,Txdx,Ty, 2.29
a >0, b, c, e≥0, abc2e≤1,φ:0,∞ → 0,∞is a lower semicontinuous functions, and
φt 0if and only ift0. Also, suppose that there existsx0 ∈ Xwithx0 Tx0. If
Tis continuous, 2.30
or
{xn} ⊂ Xis a nondecreasing sequence withxn−→zinX, then xnz∀n 2.31
holds. Then,Thas a fixed point.
Remark 2.3. InTheorem 1.122, it is proved that if
every pair of elements has a lower bound and upper bound, 2.32
then for everyx∈ X,
lim n→ ∞T
nxy, 2.33
whereyis the fixed point ofTsuch that
y lim n→ ∞T
nx
0, 2.34
and hence,Thas a unique fixed point. If condition2.32fails, it is possible to find examples of functionsT with more than one fixed point. There exist some examples to illustrate this fact in18.
Example 2.4. LetXR, and consider a relation onXas follows:
xy⇐⇒ xyorx, y∈0,1withx≤y. 2.35
It is easy to see thatis a partial order onX. Letdbe Euclidean metric onX. Now, define a self map ofXas follows:
Tx
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
2x−3
2, x >1,
x
4, 0≤x≤1, 0, x <0.
Now, we claim that the condition2.1ofTheorem 2.1is satisfied withϕt t, φt t/2. Indeed, ifx, y /∈0,1, thenxy⇔xy. Therefore, sincedTx,Ty 0, then the condition
2.1is satisfied. Again, ifx∈0,1andy /∈0,1, thenxandyare not comparative. Now, if
x, y∈0,1, thenxy⇔x≤yand
dTx,Tydx
4,
y
4
1
4d
x, y
1
2Θ
x, y,
a 1
2, bce0
Θx, y−1
2Θ
x, y
Θx, y−φΘx, y.
2.37
Also, it is easy to see that the other conditions ofTheorem 2.1are satisfied, and soThas a fixed point inX. Also, note that the weak contractive condition ofTheorem 1.3of this paper and Corollary 2.2 of7is not satisfied.
Now, we will give a common fixed point theorem for two maps. For this, we need the following definition, which is given in28.
Definition 2.5. LetX,be a partially ordered set. Two mappingsS,T:X → Xare said to be weakly increasing ifSx TSxandTx STxfor allx∈ X.
Note that two weakly increasing mappings need not be nondecreasing. There exist some examples to illustrate this fact in11.
Theorem 2.6. LetX,be a partially ordered set, and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetS,T :X → Xare two weakly increasing mappings such that
ϕdTx,Sy≤ϕΦx, y−φΦx, y, 2.38
for all comparablex, y∈ X, where
Φx, yadx, ybdx,Tx cdy,Syedy,Txdx,Sy, 2.39
a >0, b, c, e≥0, abc2e≤1,ϕ, φ:0,∞ → 0,∞, ϕis continuous, nondecreasing,φis lower semicontinuous functions, andϕt 0φtif and only if t0. If
or
Tis continuous, 2.41
or
{xn} ⊂ X is a nondecreasing sequence withxn−→z inX, thenxnz∀n 2.42
holds. Then,SandThave a common fixed point.
Remark 2.7. Note that in this theorem, we remove the condition “there exists anx0 ∈ Xwith
x0 Tx0” ofTheorem 2.1. Again, we can consider the result ofRemark 2.3for this theorem.
Proof ofTheorem 2.6. First of all we show that ifSorThas a fixed point, then it is a common fixed point ofS andT. Indeed, letzbe a fixed point ofS. Now, assumedz,Tz >0. If we use2.38forxyz, we have
ϕdTz, z ϕdTz,Sz
≤ϕΦz, z−φΦz, z
≤ϕdTz, z−φbedTz, z,
2.43
which is a contradiction. Thus,dz,Tz 0, and sozis a common fixed point ofS andT. Similarly, ifzis a fixed point ofT, then it is also fixed point ofS. Now, letx0be an arbitrary point ofX. Ifx0Sx0, the proof is finished, so assume thatx0/Sx0. We can define a sequence
{xn}inXas follows:
x2n1Sx2n, x2n2Tx2n1 forn∈ {0,1, . . .}. 2.44
Without lost of generality, we can suppose that the successive term of {xn} are different. Otherwise, we are again finished. Note that sinceSandTare weakly increasing, we have
x1 Sx0 TSx0Tx1x2,
x2 Tx1 STx1Sx2x3,
2.45
and continuing this process, we have
x1x2 · · · xnxn1 · · ·. 2.46
Now, sincex2n−1 and x2n are comparable, then we can use 2.38for these points, then we have
where
Φx2n−1, x2n adx2n−1, x2n bdx2n−1,Tx2n−1 cdx2n,Sx2n
edx2n,Tx2n−1 dx2n−1,Sx2n
≤abedx2n−1, x2n cedx2n, x2n1.
2.48
Now, ifdx2n1, x2n> dx2n, x2n−1for somen∈ {1,2, . . .}, then
Φx2n−1, x2n≤abc2edx2n1, x2n≤dx2n1, x2n, 2.49
and so, from2.47we have
ϕdx2n, x2n1≤ϕdx2n1, x2n−φΦx2n−1, x2n, 2.50
which is a contradiction. So, we have dx2n1, x2n ≤ dx2n, x2n−1 for all n ∈ {1,2, . . .}. Similarly, we havedx2n1, x2n2 ≤ dx2n, x2n1 for alln ∈ {0,1, . . .}. Therefore, we have for alln∈ {1,2, . . .}
dxn1, xn≤dxn, xn−1, 2.51
and so the sequence {dxn1, xn} is nonincreasing and bounded below. Thus, there exists ρ ≥ 0 such that limn→ ∞dxn1, xn ρ. This implies that limn→ ∞dx2n, x2n1 limn→ ∞dx2n−1, x2n ρ. Suppose thatρ >0. Therefore, from2.39,
lim sup n→ ∞
adx2n−1, x2n≤lim sup n→ ∞ Φ
x2n−1, x2n
≤lim sup
n→ ∞ {abedx2n−1, x2n cedx2n, x2n1}.
2.52
This implies 0 < aρ ≤ lim supn→ ∞Φx2n−1, x2n ≤ a b c 2eρ ≤ ρ, and so there exist ρ1 > 0 and a subsequence {Φx2nk−1, x2nk} of {Φx2n−1, x2n} such that limk→ ∞Φx2nk−1, x2nk ρ1 ≤ρ.
By the lower semicontinuity ofφ, we have
φρ1
≤lim inf k→ ∞ φ
Φx2nk−1, x2nk. 2.53
Now, from2.38, we have
ϕdx2nk, x2nk1
ϕdTx2nk−1,Sx2nk
≤ϕΦx2nk−1, x2nk−φΦx2nk−1, x2nk,
and taking upper limit ask → ∞, we have
ϕρ≤ϕρ1
−lim inf k→ ∞ φ
Φx2nk−1, x2nk
≤ϕρ1
−φρ1
≤ϕρ−φρ1
,
2.55
which is a contradiction. Therefore, we have
ρ0 lim
n→ ∞dxn1, xn. 2.56
Next, we show that{xn} is a Cauchy sequence. For this, it is sufficient to show that
{x2n}is a Cauchy sequence. Suppose it is not true. Then, we can find anδ > 0 such that for each even integer 2k, there exist even integers 2mk>2nk>2ksuch that
dx2nk, x2mk
≥δ fork∈ {1,2, . . .}. 2.57
We may also assume that
dx2mk−2, x2nk< δ, 2.58
by choosing 2mk to be smallest number exceeding 2nk for which 2.57 holds. Now,
2.56,2.57, and2.58imply
0< δ≤dx2nk, x2mk
≤dx2nk, x2mk−2
dx2mk−2, x2mk−1
dx2mk−1, x2mk
≤δdx2mk−2, x2mk−1
dx2mk−1, x2mk,
2.59
and so
lim k→ ∞d
x2nk, x2mkδ. 2.60
Also, by the triangular inequality,
dx2nk, x2mk−1
−dx2nk, x2mk≤dx2mk−1, x2mk,
dx2nk1, x2mk−1
−dx2nk, x2mk≤d
x2mk−1, x2mk
dx2nk, x2nk1
. 2.61
Therefore, we get
lim k→ ∞d
x2nk, x2mk−1
δ,
lim k→ ∞d
x2nk1, x2mk−1
δ.
On the other hand, sincex2nkandx2mk−1are comparable, we can use the condition2.38 for these points. Since
lim k→ ∞Φ
x2mk−1, x2nk lim
k→ ∞ ad
x2mk−1, x2nkbdx2mk−1,Tx2mk−1
cdx2nk,Sx2nk
edx2nk,Tx2mk−1
dx2mk−1,Sx2nk
lim k→ ∞ ad
x2mk−1, x2nk
bdx2mk−1, x2mk
cdx2nk, x2nk1
edx2nk, x2mk
dx2mk−1, x2nk1
a2eδ,
2.63
we have
ϕδ≤lim sup k→ ∞ ϕ
dx2nk, x2mk
≤lim sup k→ ∞
ϕdx2nk, x2nk1
dx2nk1, x2mk
≤lim sup k→ ∞ ϕ
dx2nk, x2nk1
dSx2nk,Tx2mk−1
lim sup k→ ∞
ϕdSx2nk,Tx2mk−1
≤lim sup k→ ∞
ϕΦx2mk−1, x2nk−φΦx2mk−1, x2nk
ϕa2eδ−lim inf k→ ∞ φ
Φx2mk−1, x2nk
≤ϕa2eδ−φa2eδ
≤ϕδ−φa2eδ.
2.64
This is a contradiction. Thus,{x2n}is a Cauchy sequence inX, so{xn}is a Cauchy sequence. Therefore, there exists az∈ Xwith limn→ ∞xnz.
IfSorTis continuous hold, then clearly,zSzTz. Now, suppose that2.42holds anddSz, z>0. Since limn→ ∞xn z, then from2.42,x2n−1 zfor alln. Using2.38, we have
ϕdTx2n−1,Sz≤ϕΦx2n−1, z−φΦx2n−1, z, 2.65
or
so taking upper limit from the last inequality, we have
ϕdz,Sz≤ϕcedz,Sz−φcedz,Sz, 2.67
which is a contradiction. Thus,dz,Sz 0, and sozSzTz.
Corollary 2.8. LetX,be a partially ordered set, and suppose that there exists a metricdin X such thatX, dis a complete metric space. LetS,T :X → Xbe two weakly increasing mappings such that
ϕdTx,Sy≤ϕdx, y−φdx, y, 2.68
for all comparablex, y ∈ X, whereϕ, φ:0,∞ → 0,∞, ϕis a continuous, nondecreasing,φis lower semicontinuous functions, andϕt 0φtif and only ift0. If
S is continuous, 2.69
or
Tis continuous, 2.70
or
{xn} ⊂ Xis a nondecreasing sequence withxn−→zinX, then xnz∀n 2.71
holds. Then,SandThave a common fixed point.
Corollary 2.9. LetX,be a partially ordered set, and suppose that there exists a metricdin X such thatX, dis a complete metric space. LetS,T:X → Xbe two weakly increasing mappings such that
dTx,Sy≤Φx, y−φΦx, y, 2.72
for all comparablex, y∈ X, where
Φx, yadx, ybdx,Tx cdy,Syedy,Txdx,Sy, 2.73
a >0, b, c, e≥0, abc2e≤1,φ:0,∞ → 0,∞is a lower semicontinuous functions, and
φt 0if and only ift0. If
S is continuous, 2.74
or
or
{xn} ⊂ Xis a nondecreasing sequence withxn−→zinX, then xnz∀n 2.76
holds. Then,SandThave a common fixed point.
3. Some Applications
In this section, we present some applications of previous sections, first we obtain some fixed point theorems for single mapping and pair of mappings satisfying a general contractive condition of integral type in complete partially ordered metric spaces. Second, we give an existence theorem for common solution of two integral equations.
SetΥ {Ψ:R → R :Ψis a Lebesgue integrable mapping which is summable and nonnegative and satisfies0Ψtdt >0, for each >0}.
Theorem 3.1. LetX,be a partially ordered set and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetT:X → Xbe a nondecreasing mapping such that
ϕdTx,Ty
0
Ψtdt≤
ϕΘx,y
0
Ψtdt−
φΘx,y
0
Ψtdt foryx 3.1
where
Θx, yadx, ybdx,Tx cdy,Tyedy,Txdx,Ty 3.2
a >0, b, c, e ≥0, abc2e ≤ 1,ϕ, φ:0,∞ → 0,∞, ϕis continuous, nondecreasing,φ
is lower semicontinuous functions, andϕt 0φtif and only ift0. Also, suppose that there existsx0 ∈ Xwithx0 Tx0. If
Tis continuous, 3.3
or
{xn} ⊂ X is a nondecreasing sequence withxn−→zinX, thenxnz ∀n 3.4
holds. Then,Thas a fixed point.
Proof. DefineΛ : R → R byΛx 0xΨtdt, then Λis continuous and nondecreasing withΛ0 0. Thus,3.1becomes
which further can be written as
ϕ1
dTx,Ty≤ϕ1
Θx, y−φ1
Θx, y, 3.6
whereϕ1 Λ◦ϕandφ1 Λ◦φ. Hence byTheorem 2.1has unique fixed fixed point.
Theorem 3.2. LetX,be a partially ordered set, and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetS,T:X → Xbe two weakly increasing mappings such that
ϕdTx,Sy
0 Ψ
tdt≤
ϕΘx,y
0 Ψ
tdt−
φΘx,y
0 Ψ
tdt foryx, 3.7
for all comparablex, y∈ X, where
Φx, yadx, ybdx,Tx cdy,Syedy,Txdx,Sy, 3.8
a >0, b, c, e≥0, abc2e≤1,ϕ, φ:0,∞ → 0,∞, ϕis continuous, nondecreasing,φis lower semicontinuous functions, andϕt 0φtif and only ift0. If
S is continuous, 3.9
or
Tis continuous, 3.10
or
{xn} ⊂ Xis a nondecreasing sequence withxn−→zinX, then xnz∀n 3.11
holds. Then,SandThave a common fixed point.
Proof. DefineΛ : R → R byΛx 0xΨtdt, then Λis continuous and nondecreasing withΛ0 0. Thus,3.7becomes
ΛϕdTx,Sy≤ΛϕΘx, y−ΛφΘx, y, 3.12
which further can be written as
ψ1
dTx,Sy≤ψ1
Θx, y−ϕ1
Θx, y, 3.13
Now, consider the integral equations
xt b
a
K1t, s, xsdsgt, t∈a, b,
xt b
a
K2t, s, xsdsgt, t∈a, b.
3.14
Letbe a partial order relation onRn.
Theorem 3.3. Consider the integral equations3.14.
iK1, K2:a, b×a, b×Rn → Rnandg:Rn → Rn are continuous,
iifor eacht, s∈a, b,
K1t, s, xsK2
t, s,
b
a
K1s, τ, xτdτgs
,
K2t, s, xsK1
t, s,
b
a
K2s, τ, xτdτgs
,
3.15
iiithere exist a continuous functionp:a, b×a, b → Rsuch that
|K1t, s, u−K2t, s, v| ≤pt, s
ln|u−v|21 3.16
for eacht, s∈a, band comparableu, v∈Rn,
ivsupt∈a,babpt, s2ds≤1/b−a.
Then, the integral equations3.14have a unique common solutionx∗inCa, b,Rn.
Proof. LetX:Ca, b,Rnwith the usual supremum norm; that is,xmax
t∈a,b|xt|, for
x∈Ca, b,Rn. Consider onXthe partial order defined by
x, y∈Ca, b,Rn, xyiffxtytfor anyt∈a, b. 3.17
Then,X,is a partially ordered set. Also,X, · is a complete metric space. Moreover, for any increasing sequence{xn}inX converging tox∗ ∈ X, we havexnt x∗tfor any
t ∈a, b. Also, for everyx, y ∈ X, there existscx, y∈ Xwhich is comparable toxandy
DefineT, S:X → X, by
Txt b
a
K1t, s, xsdsgt, t∈a, b,
Sxt b
a
K2t, s, xsdsgt, t∈a, b.
3.18
Now, fromii, we have for allt∈a, b,
Txt b
a
K1t, s, xsdsgt
b a K2 t, s, b a
K1s, τ, xτdτgs
dsgt
b
a
K2t, s, Txsdsgt
STxt,
Sxt b
a
K2t, s, xsdsgt
b a K1 t, s, b a
K2s, τ, xτdτgs
dsgt
b
a
K1t, s, Sxsdsgt
TSxt.
3.19
Thus, we haveTx STxandSx TSxfor allx∈ X. This shows thatTandSare weakly increasing. Also, for each comparablex, y∈ X, we have
Txt− Syt
b
a
K1t, s, xsds−
b
a
K2
t, s, ysds
≤
b
a
K1t, s, xs−K2
t, s, ysds
≤
b
a
pt, s
lnxs−ys21ds
≤
b
a
pt, s2ds
1/2b
a
lnxs−ys21ds 1/2
.
Hence,
Tx− Sy2≤ sup t∈a,b
b
a
pt, s2ds b
a
lnxs−ys21ds
≤lnx−y21
x−y2−x−y2−lnx−y21.
3.21
Putϕx x2, φx x2−lnx21. Therefore,
ϕTx− Sy≤ϕx−y−φx−y, 3.22
for each comparablex, y ∈ X. Therefore, all conditions ofCorollary 2.8are satisfied. Thus, the conclusion follows.
Acknowledgments
The authors thank the referees for their appreciation, valuable comments, and suggestions.
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