Volume 2011, Article ID 380784,14pages doi:10.1155/2011/380784
Research Article
Contractive Mapping in Generalized,
Ordered Metric Spaces with Application in Integral Equations
L. Gholizadeh,
1R. Saadati,
1W. Shatanawi,
2and S. M. Vaezpour
31Department of Mathematics, Science and Research Branch, Islamic Azad University (IAU), Tehran, Iran
2Department of Mathematics, Hashemite University, P.O. Box 150459, Zarqa 13115, Jordan
3Department of Mathematics, Amirkabir University of Technology, Tehran, Iran
Correspondence should be addressed to R. Saadati,[email protected] Received 22 June 2011; Revised 1 October 2011; Accepted 3 October 2011 Academic Editor: Cristian Toma
Copyrightq2011 L. Gholizadeh 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 consider the concept ofΩ-distance on a complete, partially orderedG-metric space and prove some fixed point theorems. Then, we present some applications in integral equations of our obtained results.
1. Introduction
The Banach fixed point theorem for contraction mapping has been generalized and extended in many directions 1–11. Nieto and Rodr´ıguez-L ´opez 10, Ran and Reurings 12, and Petrusel and Rus13presented some new results for contractions in partially ordered metric spaces. The main idea in10,12,14involves combining the ideas of an iterative technique in the contraction mapping principle with those in the monotone technique. Also, Mustafa and Sims15introduced the concept ofG-metric. Some authors16,17have proved some fixed point theorems in these spaces. Recently, Saadati et al. 18, using the concept of G- metric, defined an Ω-distance on completeG-metric space and generalized the concept of w-distance due to Kada et al.19.
In this paper, we extend some recent fixed point theorems by using this concept and prove various fixed point theorems in generalized partially orderedG-metric spaces.
At first we recall some definitions and lemmas. For more information see15–18,20–
23.
Definition 1see15. LetXbe a nonempty set. A functionG:X×X×X → 0,∞is called aG-metric if the following conditions are satisfied:
iGx, y, z 0 ifxyzcoincidence, iiGx, x, y>0 for allx, y∈X, wherex /y,
iiiGx, x, z≤Gx, y, zfor allx, y, z∈X, withz /y,
ivGx, y, z Gp{x, y, z}, wherepis a permutation ofx, y, zsymmetry, vGx, y, z≤Gx, a, a Ga, y, zfor allx, y, z, a∈Xrectangle inequality.
AG-metric is said to be symmetric ifGx, y, y Gy, x, xfor allx, y∈X.
Definition 2. LetX, Gbe aG-metric space,
1a sequence{xn}inXis said to beG-Cauchy sequence if, for eachε >0, there exists a positive integern0such that for allm, n, l≥n0,Gxn, xm, xl< ε;
2a sequence{xn}inXis said to beG-convergent to a pointx∈X if, for eachε >0, there exists a positive integern0such that for allm, n≥n0,Gxm, xn, x< ε.
Definition 3see15. LetX, Gbe aG-metric space. Then a functionΩ:X × X × X → 0,∞is called anΩ-distance onXif the following conditions are satisfied:
a Ωx, y, z≤Ωx, a, a Ωa, y, zfor allx, y, z, a∈X,
bfor anyx, y∈X, Ωx, y,·,Ωx,·, y:X → 0,∞are lower semicontinuous, cfor eachε >0, there exists aδ >0 such thatΩx, a, a≤δandΩa, y, z≤δimply
Gx, y, z≤ε.
Example 1see18. LetX, dbe a metric space andG:X3 → 0,∞defined by G
x, y, z
max d
x, y , d
y, z
, dx, z
, 1.1
for allx, y, z∈X. ThenΩ Gis anΩ-distance onX.
Example 2see18. InX Rwe consider theG-metricGdefined by
G x, y, z
1
3x−yy−z|x−z|
, 1.2
for allx, y, z∈R. ThenΩ:R3 → 0,∞defined by Ω
x, y, z 1
3
|z−x|x−y, 1.3
for allx, y, z∈Ris anΩ-distance onR.
For more example see18.
Lemma 1.1see18. LetX be a metric space with metricG andΩbe anΩ-distance onX. Let xn, yn be sequences inX, αn, βn be sequences in0,∞converging to zero and letx, y, z, a ∈ X.
Then one has the following.
1IfΩy, xn, xn≤αnandΩxn, y, z≤βnforn∈N, thenGy, y, z< εand henceyz.
2IfΩyn, xn, xn≤αnandΩxn, ym, z≤βnform > nthenGyn, ym, z → 0 and hence yn → z.
3IfΩxn, xm, xl≤αnfor anyl, m, n∈Nwithn≤m≤l, thenxnis a G-Cauchy sequence.
4IfΩxn, a, a≤αnfor anyn∈Nthenxnis a G-Cauchy sequence.
Definition 4see18. G-metric spaceXis said to beΩ-bounded if there is a constantM >0 such thatΩx, y, z≤Mfor allx, y, z∈X.
2. Fixed Point Theorems on Partially Ordered G-Metric Spaces
Definition 5. SupposeX, ≤is a partially ordered space andT :X → Xis a mapping ofX into itself. We say thatT is nondecreasing if forx, y∈X,
x≤y⇒Tx≤T y
. 2.1
Theorem 2.1. LetX, ≤be a partially ordered space. Suppose that there exists aG-metric onXsuch thatX, Gis a completeG-metric space andΩis anΩ-distance onXsuch thatXisΩ-bounded. Let f:X → Xandg:X → Xweakly compatible andf, gbe non-decreasing mapping such that
agX⊆fX;
b Ωgx, gy, gz≤kmax{Ωfx, fy, fz,Ωfx, gx, fz,Ωfy, gy, fz,Ωfx, gy, fz, Ωfy, gx, fz}; for allx, y, z∈X and 0≤k <1,
cfor every x ∈ X and y ∈ X with fy/gy, inf{Ωfx, y, fx Ωfx, y, gx Ωfx, gx, y:fx≤gx}>0;
dthere existx0 ∈Xthatfx0≤gx0; thenf andghave a unique common fixed pointu inXandΩu, u, u 0.
Proof. Letx0 ∈ X thatfx0 ≤ gx0. By parta, we can choosex1 ∈ X such thatfx1
gx0. Again from parta, we can choosex2 ∈ X such thatfx2 gx1. Continuing this process we can construct sequences{xn}inXsuch that,
yngxnfxn1, ∀n≥0,
xn≤xn1. 2.2
Now, sincegis non-decreasing mapping then,
gxn≤gxn1, ∀ n≥0, 2.3
so, for alls≥0, Ω
yn, yn1, yns Ω
gxn, gxn1, gxns
≤kmax Ω
fxn, fxn1, fxns ,Ω
fxn, gxn, fxns ,Ω
fxn1, gxn1, fxns , Ω
fxn, gxn1, fxns ,Ω
fxn1, gxn, fxns kmax
Ω
yn−1, yn, yns−1 ,Ω
yn−1, yn, yns−1 ,Ω
yn, yn1, yns−1 , Ω
yn−1, yn1, yns−1 ,Ω
yn, yn, yns−1 .
2.4 Then,
Ω
yn, yn1, yns
≤kmax Ω
yn−1, yn, yns−1 ,Ω
yn, yn1, yns−1 , Ω
yn−1, yn1, yns−1 ,Ω
yn, yn, yns−1
. 2.5
Now since, Ω
yn−1, yn1, yns−1
≤kmax Ω
yn−2, yn, yns−2 ,Ω
yn−2, yn−1, yns−2 ,Ω
yn, yn1, yns−2 , Ω
yn−2, yn1, yns−2 ,Ω
yn, yn−1, yns−2 Ω
yn, yn, yns−1
≤kmax Ω
yn−1, yn−1, yns−2 ,Ω
yn−1, yn, yns−2 ,Ω
yn−1, yn, yns−2 , Ω
yn−1, yn, yns−2 ,Ω
yn−1, yn, yns−2 ,
2.6
thus, Ω
yn, yn1, yns
≤k2max Ω
yi, yj, yt
, n−2≤i≤n, n−1≤j≤n1, ns−2≤t≤ns−1 ...
≤kn−1max Ω
yi, yj, yt
; 1≤i≤n,2≤j ≤n1, s1≤t≤ns−1 . 2.7
SoΩyn, yn1, yns≤kn−1Mn,swhere Mn,s:max
Ω
yi, yj, yt
, 1≤i≤n,2≤j ≤n1, s1≤t≤ns−1
. 2.8
Now, for anyl > m > nwithmnkandlmtk, t∈N, we have,
m,n,llim→ ∞Ω
yn, ym, yl
0. 2.9
SinceXisΩ-bounded and Ω
yn, ym, yl
≤Ω
yn, yn1, yn1 Ω
yn1, ym, yl
≤Ω
yn, yn1, yn1 Ω
yn1, yn2, yn2
· · · Ω
ym−1, ym, yl
≤kn−1Mn,1knMn1,2· · ·km−2Mm−1,t1
≤n−m2
j1
kn−jM≤ kn−1 1−kM,
2.10
so, by Part 3 of Lemma 1.1,{yn} is a G-Cauchy sequence. Since X is G-complete, {yn} converges to a pointy∈X. Thus, forε >0 and by the lower semicontinuity ofΩ, we have
Ω
yn, ym, y
≤lim inf
p→ ∞ Ω
yn, ym, yp
≤ε, m≥n Ω
yn, y, yl
≤lim inf
p→ ∞ Ω
yn, yp, yl
≤ε, l≥n. 2.11
Assume thatfy /gy. Since,
ynfxn1gxn≤gxn1 fxn2yn1, 2.12
so,yn≤yn1, and, 0<inf
Ω
yn, y, yn Ω
yn, yn1, y Ω
yn, y, yn1
≤3ε, 2.13
for everyε >0, that is a contraction. So, we havefygy. Then, byb, Ω
gy, gy, gy
≤kΩ
gy, gy, gy
, 2.14
so,Ωgy, gy, gy 0. Similarly,Ωg2y, g2y, gy 0.
Now,
Ω
gy, g2y, gy ≤kmax Ω
gy, g2y, gy ,Ω
g2y, gy, gy , Ω
g2y, g2y, gy ,Ω
gy, gy, gy kmax
Ω
gy, g2y, gy ,Ω
g2y, gy, gy Ω
g2y, gy, gy ≤kmax Ω
gy, g2y, gy ,Ω
g2y, gy, gy .
2.15
Thus,
Ω
gy, g2y, gy 0, Ω
g2y, gy, gy 0. 2.16
By Partc ofDefinition 3,Gg2y, g2y, gy 0 and consequentlyg2y gywhich implies thatgyis a fixed point forg. Now,
f gy
g fy
g2ygy. 2.17
So, it is enough to putgyu, thenuis a common fixed point offandg.
Uniqueness: Assume that there existv∈Xsuch thatfvgvv. Hence, we have,
Ωv, v, v≤kΩv, v, v, 2.18
and soΩv, v, v 0. Also,Ωv, v, u 0. On the other hand, Ωv, u, u≤kmax{Ωv, u, u,Ωu, v, u},
Ωu, v, u≤kmax{Ωu, v, u,Ωv, u, u}, 2.19 which follows that,Ωv, u, u Ωu, v, u 0. Then by Part cofDefinition 3,u vand Ωu, u, u 0.
The following corollary is a generalization of24, Theorem 2.1.
Corollary 2.2. LetX, ≤be a partially ordered space. Suppose that there exists aG-metric on X such thatX, Gis aG-metric space andΩis anΩ-distance onX such thatX beΩ-bounded. Let f:X → Xandg:X → Xbe weakly compatible andf, gbe a non-decreasing mapping such that
agX⊆fXand eitherfXorgXis complete;
bfor allx, y, z∈Xand 0≤k <1,Ωgx, gy, gz≤kΩfx, fy, fz;
cfor every x ∈ X and y ∈ X with fy/gy, inf{Ωfx, y, fx Ωfx, y, gx Ωfx, gx, y:fx≤gx}>0;
dthere existx0∈Xthatfx0≤gx0;
thenfandghave a unique common fixed pointyinXandΩy, y, y 0.
Definition 6see25. LetΦbe the set of all functionsϕsuch thatϕ :0,∞ → 0,∞is a continuous and nondecreasing function withϕt < tfor allt ∈R and∞
n1ϕnt < ∞for eacht∈R. The functionϕis called a growth or control function ofT:X → X.
It is clear that
limn→ ∞ϕnt 0, ∀t∈R, ϕn0 0. 2.20 Theorem 2.3. LetX, ≤be a partially ordered space. Suppose that there exists aG-metric onX such thatX, Gis a completeG-metric space andΩis anΩ-distance onXandTis a non-decreasing mapping fromXinto itself. LetXbeΩ-bounded. Suppose thatϕ∈Φand
Ω
Tx, T2x, Tw ≤ϕΩx, Tx, w ∀x≤Tx, w∈X. 2.21
Also, for everyx∈X
inf Ω
x, y, x Ω
x, y, Tx Ω
x, T2x, y :x≤Tx
>0, 2.22
for everyy∈Xwithy /Ty. If there exists anx0∈Xwithx0≤Tx0, thenThas a unique fixed point.
Moreover, ifvTv, thenΩv, v, v 0.
Proof. Ifx0 Tx0, then the proof is finished. Suppose thatTx0/x0. sincex0 ≤ Tx0 andT is non-decreasing, we obtain
x0≤Tx0≤T2x0≤ · · · ≤Tn1x0≤ · · · 2.23
For alln∈Nandt≥0, Ω
Tnx0, Tn1x0, Tntx0 ≤ϕ Ω
Tn−1x0, Tnx0, Tnt−1x0
≤ϕ2 Ω
Tn−2x0, Tn−1x0, Tnt−2x0
...
≤ϕn Ω
x0, Tx0, Ttx0
.
2.24
We claim that formnkandlmtk, t∈Nwithl > m > n,
m,n,llim→ ∞Ω
Tnx0, Tmx0, Tlx0 0. 2.25 We prove by,
Ω
Tnx0, Tmx0, Tlx0 ≤Ω
Tnx0, Tn1x0, Tn1x0 Ω
Tn1x0, Tmx0, Tlx0
≤Ω
Tnx0, Tn1x0, Tn1x0 Ω
Tn1x0, Tn2x0, Tn2x0 · · · Ω
Tm−1x0, Tmx0, Tlx0
≤ϕnΩx0, Tx0, Tx0 ϕn1Ωx0, Tx0, Tx0 · · ·ϕm−2Ωx0, Tx0, Tx0 ϕm−1
Ω
x0, Tx0, Tt1x0
≤ϕn−1M ∞
n1
ϕnM
.
2.26
Since∞
n1ϕnM<∞, so,
m,nlim→ ∞Ω
Tnx0, Tmx0, Tlx0 0. 2.27
By Part c of Lemma 1.1{Tnx0} is a G-Cauchy sequence. Since X is G-complete, {Tnx0} converges to a pointu∈X. Letn∈Nbe fixed. By lower semicontinuity ofΩ,
ΩTnx0, Tmx0, u≤lim inf
p→ ∞ ΩTnx0, Tmx0, Tpx0≤ε, m > n, Ω
Tnx0, u, Tlx0 ≤lim inf
p→ ∞ ΩTnx0, Tpx0, Tmx0≤ε, l≥n.
2.28
Assume thatu /Tu. SinceTnx0≤Tn1x0, 0<inf
ΩTnx0, u, Tnx0 Ω
Tnx0, u, Tn1x0 Ω
Tnx0, Tn2x0, u :n∈N
≤3ε, 2.29
for everyε >0, which is a contraction. Therefore, we haveuTu.
Uniqueness: letvbe another fixed point ofT, then Ωu, u, v Ω
Tu, T2u, Tv ≤ϕΩu, Tu, v<Ωu, u, v, 2.30
which is a contraction. Therefore, fixed pointuis unique. Now, ifvTv, we have, Ωv, v, v Ω
Tv, T2v, T3v ≤ϕ Ω
v, Tv, T2v
ϕΩv, v, v. 2.31
SoΩv, v, v 0.
Corollary 2.4. Let the assumptions ofTheorem 2.3hold and
Ω
Tmx, Tm1x, Tmw ≤ϕΩx,Tx, w ∀m∈N, x≤Tx, w∈X, 2.32
thenT has a unique fixed point.
Proof. FromTheorem 2.3,Tmhas a unique fixed pointu. However,
TuTTmu Tm1uTmTu, 2.33
soTuis also a fixed point ofTm. Since the fixed point ofTmis unique, it must be the case that Tuu.
Corollary 2.5. Let the assumptions ofTheorem 2.3hold andT :X → Xsatisfies, Ω
Tx, T2x, Tx ≤ϕΩx, Tx, x ∀x≤Tx. 2.34
ThenT has a unique fixed point.
Proof. Takewx, and applyTheorem 2.3.
Theorem 2.6. LetX, ≤be a partially ordered space. Suppose that there exists aG-metric onX such thatX, Gis a completeG-metric space,Ωis anΩ-distance onX, andT is a non-decreasing mapping fromXinto itself. LetXbeΩ-bounded. Suppose that
Ω
Tx, T2x, Tw ≤k Ω
x, T2x, Tw Ωx, Tx, Tx , 2.35
wherex≤Tx, w∈X, k∈0,1/3. Also for everyx∈X, inf
Ω x, y, x
Ω
x, y, Tx Ω
x, T2x, y :x≤Tx
>0, 2.36
for everyy∈Xwithy /Ty. If there exists anx0∈Xwithx0≤Tx0, thenT has a unique fixed point sayuandΩu, u, u 0.
Proof. Letx0 ∈X be an arbitrary point, and define the sequencexn byxn Tnx0. By2.35 and for allt≥0,
Ωxn, xn1, xnt≤kΩxn−1, xn1, xnt Ωxn−1, xn, xn. 2.37
But by PartaofDefinition 3,
Ωxn−1, xn1, xnt≤Ωxn−1, xn, xn Ωxn, xn1, xnt. 2.38
Hence,
Ωxn, xn1, xnt≤k2Ωxn−1, xn, xn Ωxn, xn1, xnt, 2.39
which implies,
Ωxn, xn1, xnt≤ 2k
1−kΩxn−1, xn, xn. 2.40 Letr2k/1−k, thenr <1 and by repeated application of2.40, we have
Ωxn, xn1, xnt≤rnΩx0, x1, x1. 2.41
Now, for anyl > m > nwithmnkandlmtk, t∈N, we have, Ωxn, xm, xl≤Ωxn, xn1, xn1 Ωxn1, xn2, xn2
Ωxn2, xn3, xn3 · · · Ωxm−1, xm, xl
≤
rnrn1· · ·rm−1 Ωx0, x1, x1
≤ rn
1−rΩx0, x1, x1.
2.42
So,
m,n,llim→ ∞Ωxn, xm, xl 0. 2.43
By Part3ofLemma 1.1,xnis aG-Cauchy sequence. SinceXisG-complete,xnconverges to a pointu∈X. Now, similar to provingTheorem 2.1,Thas a unique fixed point andΩu, u, u 0.
Corollary 2.7. Let the assumptions ofTheorem 2.6hold and Ω
Tmx, Tm2x, Tmw ≤k Ω
x, Tm2x, Tmw Ωx, Tmx, Tmx 2.44
wherek∈0,1/3, thenT has a unique fixed point.
Proof. The argument is similar to that used in the proof ofCorollary 2.4.
3. Applications
In this section, we give an existence theorem for a solution of a class of integral equations.
Denote by Λ the set of all functions λ : 0,∞ → 0,∞ satisfying the following hypotheses:
iλis a Lebesgue-integrable mapping on each compact of0,∞, iifor every >0, we have
0λsds >0, iiiλ<1, whereλdenotes to the norm ofλ.
Now, we have the following results.
Theorem 3.1. LetX, ≤be a partially ordered space. Suppose that there exists aG-metric onX such thatX, Gis a completeG-metric space andΩis anΩ-distance onXandTis a non-decreasing mapping fromXinto itself. LetXbeΩ-bounded. Suppose that
Ω
Tx, T2x, Tw ≤
Ωx,Tx,w
0
αsds, 3.1
whereα∈Λ. Also, suppose that for everyx∈X inf
Ω x, y, x
Ω
x, y, Tx Ω
x, T2x, y :x≤Tx
>0, 3.2
for everyy∈Xwithy /Ty. If there exists anx0∈Xwithx0≤Tx0, thenThas a unique fixed point.
Proof. Defineφ : 0,∞ → 0,∞byφt t
0αsds. It is clear thatφ is nondecreasing and continuous. Fromiii, we have
φt φt
t
0
λsds ≤
t
0
|λs|ds≤ λt < t. 3.3
Also, note that
φ2t φ
φt
≤ λφt≤ λ2t. 3.4
In general, we haveφnt≤ λnt. Thus, we have ∞
n1
φnt≤∞
n1
λnt λt
1− λ <∞. 3.5
Therefore φ satisfies all the hypotheses of Definition 6. By inequality 3.1, we have ΩTx, T2x, Tw≤φΩx, Tx, w. Therefore byTheorem 2.3,T has a unique fixed point.
Now, our aim is to give an existence theorem for a solution of the following integral equation:
ut 1
0
Kt, s, usdsgt, t∈0,1. 3.6
LetXC0,1be the set of all continuous functions defined on0,1. Define
G:X×X×X−→R 3.7
by
G x, y, z
maxx−y,x−z,y−z, 3.8 wherex sup{|xt| : t ∈ 0,1}. Then X, Gis a completeG-metric space. Let Ω G.
ThenΩis anΩ-distance onX.
Define an ordered relation≤onXby
x≤y iffxt≤yt, ∀t∈0,1. 3.9
ThenX,≤is a partially ordered set. Now, we prove the following result.
Theorem 3.2. Suppose the following hypotheses hold.
aK:0,1×0,1×R → Randg:R → Rare continuous.
bKis nondecreasing in its first coordinate andgis nondecreasing.
cThere exist a continuous functionG:0,1×0,1 → 0,∞such that
|Kt, s, u−Kt, s, v| ≤Gt, s|u−v|, 3.10
for each comparableu, v∈Rand eacht, s∈0,1.
dsupt∈0,11
0Gt, sds≤rfor somer <1.
Then the integral equation3.6has a solutionu∈C0,1.
Proof. DefineT :C0,1 → C0,1by
Txt 1
0
Kt, s, xsdsgt, t∈0,1. 3.11
By hypothesisb, we have thatTis nondecreasing.
Now, if inf
Ω x, y, x
Ω
x, y, Tx Ω
x, T2x, y :x≤Tx
0, 3.12
fory ∈C0,1withy /Ty, then for eachn ∈Nthere existsxn ∈C0,1withxn ≤Txn such that
Ω
xn, y, xn
Ω
xn, y, Txn
Ω
xn, T2xn, y ≤ 1
n. 3.13
So, we have
Ω
xn, y, Txn
maxxn−y,xn−Txn,y−Txn≤ 1
n. 3.14
Therefore, for eacht∈0,1, we have
nlim→∞xnt yt,
nlim→∞Txnt yt. 3.15
By the continuity ofK, we have
yt lim
n→∞Txnt
1
0
K
t, s, lim
n→∞xns
dsgt
1
0
K
t, s, ys
dsgt Tyt.
3.16
Thus, we haveyTy, a contradiction. Thus,
inf Ω
x, y, x Ω
x, y, Tx Ω
x, T2x, y :x≤Tx
>0. 3.17
Defineφ:0,∞ → 0,∞byφt rt. Forx∈C0, Twithx≤Tx, we have Ω
Tx, T2x, Tx sup
t∈0,1
Txt−T2xt
sup
t∈0,1
1
0
Kt, s, xs−Kt, s, Txsds
≤ sup
t∈0,1
1
0
|Kt, s, xs−Kt, s, Txs|ds
≤ sup
t∈0,1
1
0
Gt, s|xs−Txs|ds
≤ sup
t∈0,1|xt−Txt|sup
t∈0,1
T
0
Gt, sds Ωx, Tx, xsup
t∈0,1
1
0
Gt, sds
≤rΩx, Tx, x φΩx, Tx, x.
3.18
Moreover, takex0 0, thenx0 ≤ Tx0. Thus all the required hypotheses ofCorollary 2.5are satisfied. Thus there exists a solutionu∈C0, Tof the integral equation3.6.
Acknowledgment
The authors would like to thank the referee and area editor Professor Cristian Toma for providing useful suggestions and comments for the improvement of this paper.
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