Volume 2010, Article ID 621469,17pages doi:10.1155/2010/621469
Research Article
Some Fixed Point Theorems on Ordered Metric Spaces and Application
Ishak Altun and Hakan Simsek
Department of Mathematics, Faculty of Science and Arts, Kirikkale University, Yahsihan, Kirikkale 71450, Turkey
Correspondence should be addressed to Ishak Altun,[email protected] Received 2 July 2009; Accepted 13 January 2010
Academic Editor: Juan Jose Nieto
Copyrightq2010 I. Altun and H. Simsek. 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 present some fixed point results for nondecreasing and weakly increasing operators in a partially ordered metric space using implicit relations. Also we give an existence theorem for common solution of two integral equations.
1. Introduction
Existence of fixed points in partially ordered sets has been considered recently in1, and some generalizations of the result of1 are given in2–6. Also, in1some applications to matrix equations are presented, in3,4 some applications to periodic boundary value problem and to some particular problems are, respectively, given. Later, in6O’Regan and Petrus¸el gave some existence results for Fredholm and Volterra type integral equations. In some of the above works, the fixed point results are given for nondecreasing mappings.
We can order the purposes of the paper as follows.
First, we give a slight generalization of some of the results of the above papers using an implicit relation in the following way.
In1,3, the authors used the following contractive condition in their result, there existsk∈0,1such that
d
fx, fy
≤kd x, y
foryx. 1.1
Afterwards, in2, the authors used the nonlinear contractive condition, that is, d
fx, fy
≤ψ d
x, y
foryx, 1.2
whereψ :0,∞ → 0,∞is anondecreasing function with limn→ ∞ψnt 0 fort >0, instead of1.1. Also in2, the authors proved a fixed point theorem using generalized nonlinear contractive condition, that is,
d
fx, fy
≤ψ
max
d x, y
, d x, fx
, d y, fy
,1 2
d x, fy
d
y, fx 1.3
for y x, where ψ is as above. In the Section 3, we generalized the above contractive conditions using the implicit relation technique in such a way that
T d
Fx, Fy , d
x, y
, dx, Fx, d y, Fy
, d x, Fy
, d y, Fx
≤0 1.4
for y x, where T : R6 → Ris a function as given in Section 2. We can obtain various contractive conditions from1.4. For example, if we choose
Tt1, . . . , t6 t1−ψ
max
t2, t3, t4,1
2t5t6 1.5
in1.4, then, we have1.3. Similarly we can have the contractive conditions in7–9from 1.4.
In some of the above mentioned theorems, the fixed point results are given for nondecreasing mappings. Also in these theorems the following condition is used:
there existsx0∈X such thatx0fx0. 1.6 InSection 4, we give some examples such that two weakly increasing mappings need not be nondecreasing. Therefore, we give a common fixed point theorem for two weakly increasing operators in partially ordered metric spaces using implicit relation technique. Also we did not use the condition1.6in this theorem. At the end, to see the applicability of our result, we give an existence theorem for common solution of two integral equations using a result of theSection 4.
2. Implicit Relation
Implicit relations on metric spaces have been used in many articles. See for examples,10–15.
LetR denote the nonnegative real numbers, and letTbe the set of all continuous functionsT :R6 → Rsatisfying the following conditions:
T1:Tt1, . . . , t6is nonincreasing in variablest2, . . . , t6;
T2: there exists a right continuous functionf :R → R, f0 0,ft < tfort >0, such that foru≥0,
Tu, v, u, v,0, uv≤0 2.1
or
Tu, v,0,0, v, v≤0 2.2
impliesu≤fv;
T3:Tu,0, u,0,0, u>0, for allu >0.
Example 2.1. Tt1, . . . , t6 t1−αmax{t2, t3, t4}−1−αat5bt6,where 0≤α <1, 0≤a <1/2, 0≤b <1/2.
Letu > 0 andTu, v, u, v,0, uv u−αmax{u, v} −1−αbuv ≤ 0. Ifu ≥ v, then1−bu ≤ bvwhich implies b ≥ 1/2, a contradiction. Thusu < vandu ≤ α 1− αb/1−1−αbvβv. Similarly, letu >0 andTu, v,0,0, v, v u−αv−1−αabv u−α 1−αabv≤0,thenu≤α 1−αabvγv.Ifu0, thenu≤γv.ThusT2is satisfied withft max{β, γ}t. AlsoTu,0, u,0,0, u u−αu−1−αbu 1−α1−bu >0, for allu >0. Therefore,T ∈ T.
Example 2.2. Tt1, . . . , t6 t1−kmax{t2, t3, t4,1/2t5t6}, wherek∈0,1.
Letu >0 andTu, v, u, v,0, uv u−kmax{u, v} ≤0. Ifu≥v,thenu≤ku,which is a contradiction. Thusu < vandu≤kv.Similarly, letu >0 andTu, v,0,0, v, v u−kv≤ 0, then we haveu ≤ kv. If u 0, then u ≤ kv. Thus T2 is satisfied with ft kt. Also Tu,0, u,0,0, u u−ku >0,for allu >0.Therefore,T ∈ T.
Example 2.3. Tt1, . . . , t6 t1−φmax{t2, t3, t4,1/2t5t6},whereφ :R → Ris right continuous andφ0 0,φt< tfort >0.
Letu >0 andTu, v, u, v,0, uv u−φmax{u, v}≤0.Ifu≥v,thenu−φu≤0, which is a contradiction. Thusu < vandu≤φv.Similarly, letu >0 andTu, v,0,0, v, v u−φv≤0,then we haveu≤φv.Ifu0, thenu≤φv.ThusT2is satisfied withf φ.
AlsoTu,0, u,0,0, u u−φu>0, for allu >0.Therefore,T ∈ T.
Example 2.4. Tt1, . . . , t6 t21−t1at2bt3ct4−dt5t6, wherea >0,b, c, d≥0,abc <1 andad <1.
Letu >0 andTu, v, u, v,0, uv u2−uavbucv≤0.Thenu≤ac/1− bvh1v.Similarly, letu >0 andTu, v,0,0, v, v u2−auv−dv2 ≤0,then we haveu≤ a√
4da2/2vh2v.Ifu0,thenu≤h2v.ThusT2is satisfied withft max{h1, h2}t.
AlsoTu,0, u,0,0, u 1−bu2>0, for allu >0. Therefore,T ∈ T.
3. Fixed Point Theorem for Nondecreasing Mappings
We need the following lemma for the proof of our theorems.
Lemma 3.1see16. Letf:R → Rbe a right continuous function such thatft< tfor every t >0, then limn→ ∞fnt 0, wherefndenotes then-times repeated composition offwith itself.
Theorem 3.2. LetX,be a partially ordered set and suppose that there is a metricdonXsuch that X, dis a complete metric space. SupposeF :X → Xis a nondecreasing mapping such that for all x, y∈Xwithyx,
T d
Fx, Fy , d
x, y
, dx, Fx, d y, Fy
, d x, Fy
, d y, Fx
≤0, 3.1
whereT ∈ T. Also
F is continuous, 3.2
or
if {xn} ⊂X is a nondecreasing sequence withxn−→xinX,
thenxnx ∀n 3.3
hold. If there exists anx0∈Xwithx0Fx0, thenFhas a fixed point.
Proof. IfFx0 x0, then the proof is finished; so suppose x0/Fx0. Now letxn Fxn−1 for n∈ {1,2, . . .}. Notice that, sincex0Fx0andFis nondecreasing, we have
x0 x1x2 · · · xnxn1 · · ·. 3.4
Now sincexn−1xn, we can use the inequality3.1for these points, then we have TdFxn, Fxn−1, dxn, xn−1, dxn, Fxn, dxn−1, Fxn−1, dxn, Fxn−1, dxn−1, Fxn≤0
3.5 and so
Tdxn1, xn, dxn, xn−1, dxn, xn1, dxn−1, xn,0, dxn−1, xn1≤0. 3.6
Now usingT1, we have
Tdxn1, xn, dxn, xn−1, dxn, xn1, dxn−1, xn,0, dxn−1, xn dxn, xn1≤0, 3.7 and fromT2 there exists a right continuous function f : R → R,f0 0,ft < t, for t >0, such that for alln∈ {1,2, . . .},
dxn1, xn≤fdxn, xn−1. 3.8
If we continue this procedure, we can have
dxn1, xn≤fndx1, x0, 3.9
and so fromLemma 3.1,
nlim→ ∞dxn1, xn 0. 3.10
Next we show that{xn}is a Cauchy sequence. Suppose it is not true. Then we can find aδ >0 and two sequence of integers{mk},{nk}, mk> nk≥kwith
rkd
xnk, xmk
≥δ fork∈ {1,2, . . .}. 3.11
We may also assume
d
xmk−1, xnk
< δ 3.12
by choosing mkto be the smallest number exceeding nkfor which 3.11 holds. Now 3.9,3.11, and3.12imply
δ≤rk≤d
xmk, xmk−1 d
xmk−1, xnk
≤fmk−1dx0, x1 δ 3.13
and so
klim→ ∞rkδ. 3.14
Also since
δ≤rk≤d
xnk, xnk1 d
xmk, xmk1 d
xnk1, xmk1
, 3.15
we have from3.9that
δ≤rk≤fnkdx0, x1 fmkdx0, x1 d
xmk1, xnk1
. 3.16
On the other hand, since xnk xmk, we can use the condition 3.1 for these points.
Therefore, we have T
d
Fxmk, Fxnk , d
xmk, xnk , d
xmk, Fxmk , d
xnk, Fxnk , d
xmk, Fxnk , d
xnk, Fxmk
≤0 3.17
and so
T d
Fxmk, Fxnk
, rk, fmkdx0, x1, fnkdx0, x1, rkfnkdx0, x1, rkfmkdx0, x1
≤0.
3.18
Now lettingk → ∞and using3.14, we have, by continuity ofT,that T
k→ ∞limd
xmk1, xnk1
, δ,0,0, δ, δ ≤0. 3.19
FromT2, we have limk→ ∞dxmk1, xnk1 ≤ fδ. Therefore, lettingk → ∞in3.16, we haveδ≤fδ.This is a contradiction sinceft< tfort >0.Thus{xn}is a Cauchy sequence inX,so there exists anx∈Xwith limn→ ∞xnx.
If3.2holds, then clearlyx Fx. Now suppose 3.3holds. Supposedx, Fx > 0.
Now since limn→ ∞xn x, then from3.3,xn xfor alln. Using the inequality3.1, we have
TdFx, Fxn, dx, xn, dx, Fx, dxn, Fxn, dx, Fxn, dxn, Fx≤0, 3.20
so lettingn → ∞from the last inequality, we have
TdFx, x,0, dx, Fx,0,0, dx, Fx≤0, 3.21 which is a contradiction toT3. Thusdx, Fx 0 and soxFx.
Remark 3.3. Note that if we take that
T4: there exists a nondecreasing functionf :R → Rwith limn→ ∞fnt 0 for each t >0,such that foru≥0,
Tu, v, u, v,0, uv≤0 3.22 or
Tu, v,0,0, v, v≤0 3.23
impliesu≤fv,
Instead ofT2inTheorem 3.2, again we can have the same result.
If we combineTheorem 3.2withExample 2.1, we obtain the following result.
Corollary 3.4. LetX,be a partially ordered set and suppose that there is a metricdonX such thatX, dis a complete metric space. SupposeF:X → Xis a nondecreasing mapping such that for allx, y∈Xwithyx,
d
Fx, Fy
≤αmax d
x, y
, dx, Fx, d y, Fy
1−α ad
x, Fy bd
y, Fx
, 3.24 where 0≤α <1, 0≤a <1/2, 0≤b <1/2. Also
F is continuous 3.25
or
if {xn} ⊂X is a nondecreasing sequence withxn−→xinX,
thenxnx ∀n 3.26
hold. If there exists anx0∈Xwithx0Fx0, thenFhas a fixed point.
Remark 3.5. Theorem 2.2 of2follows fromExample 2.3,Remark 3.3, andTheorem 3.2.
Remark 3.6. We can have some new results from other examples andTheorem 3.2.
Remark 3.7. In Theorem 11, it is proved that if
every pair of elements has a lower bound and an upper bound, 3.27 then for everyx∈X,
nlim→ ∞Fnx y, 3.28
whereyis the fixed point ofFsuch that y lim
n→ ∞Fnx0 3.29
and henceFhas a unique fixed point. If condition3.27fails, it is possible to find examples of functionsF with more than one fixed point. There exist some examples to illustrate this fact in3.
4. Fixed Point Theorem for Weakly Increasing Mappings
Now we give a fixed point theorem for two weakly increasing mappings in ordered metric spaces using an implicit relation. Before this, we will define an implicit relation for the contractive condition of the theorem.
LetTbe the set of all continuous functionsT :R6 → RsatisfyingT1and the following conditions:
T2: there exists a right continuous functionf :R → R,f0 0, ft < tfort >0, such that foru≥0,
Tu, v, u, v,0, uv≤0 4.1 or
Tu, v, v, u, uv,0≤0 4.2 or
Tu, v,0,0, v, v≤0 4.3
impliesu≤fv;
T3:Tu,0, u,0,0, u>0 andTu,0,0, u, u,0>0, for allu >0.
We can easily show that, all functions in the Examples inSection 2are inT.
Definition 4.1see17,18. LetX,be a partially ordered set. Two mappingsF, G:X → X are said to be weakly increasing ifFxGFxandGxFGxfor allx∈X.
Note that, two weakly increasing mappings need not be nondecreasing.
Example 4.2. LetXRendowed with usual ordering. LetF, G:X → Xdefined by
Fx
⎧⎨
⎩
x if 0≤x≤1
0 if 1< x <∞, Gx
⎧⎨
⎩
√x if 0≤x≤1
0 if 1< x <∞, 4.4
then it is obvious thatFx ≤ GFxand Gx ≤ FGx for allx ∈ X. ThusF and Gare weakly increasing mappings. Note that bothFandGare not nondecreasing.
Example 4.3. LetX 1,∞×1,∞be endowed with the coordinate ordering, that is,x, y z, w⇔x≤zandy ≤w. LetF, G:X → Xbe defined byFx, y 2x,3yandGx, y x2, y2, thenFx, y 2x,3yGFx, y G2x,3y 4x2,9y2andGx, y x2, y2 FGx, y Fx2, y2 2x2,3y2. ThusFandGare weakly increasing mappings.
Example 4.4. Let X R2 be endowed with the lexicographical ordering, that is, x, y z, w⇔x < zorifxz,theny≤w. LetF, G:X → Xbe defined by
F x, y
max
x, y ,min
x, y , G
x, y
max
x, y ,xy
2 , 4.5
then
F x, y
max
x, y ,min
x, y GF
x, y G
max x, y
,min x, y
max
max x, y
,min x, y
,max x, y
min x, y 2
max x, y
,xy
2 ,
G x, y
max
x, y ,xy
2 FG
x, y F
max
x, y ,xy
2
max
max
x, y ,xy
2
,min
max x, y
,xy 2
max
x, y ,xy
2 .
4.6 Thus F and G are weakly increasing mappings. Note that 1,4 2,3 but F1,4 4,13,2 F2,3, thenFis not nondecreasing. Similarly,Gis not nondecreasing.
Theorem 4.5. LetX,be a partially ordered set and suppose that there is a metricdonXsuch that X, dis a complete metric space. SupposeF, G:X → Xare two weakly increasing mappings such that for all comparablex, y∈X,
T d
Fx, Gy , d
x, y
, dx, Fx, d y, Gy
, d x, Gy
, d y, Fx
≤0, 4.7
whereT ∈ T. Also
F is continuous 4.8
or
G is continuous 4.9
or
if {xn} ⊂X is a nondecreasing sequence withxn−→xinX,
thenxnx ∀n 4.10
hold, thenFandGhave a common fixed point.
Remark 4.6. Note that, in this theorem we remove the condition “there exists anx0 ∈Xwith x0Fx0” ofTheorem 3.2. Again we can consider the result ofRemark 3.7for this theorem.
Proof ofTheorem 4.5. First of all we show that ifForGhas a fixed point, then it is a common fixed point ofF andG. Indeed, letzbe a fixed point ofF. Now assumedz, Gz > 0. If we use the inequality4.7, forxyz, we have
Tdz, Gz,0,0, dz, Gz, dz, Gz,0≤0, 4.11
which is a contradiction toT3. Thusdz, Gz 0 and sozis a common fixed point ofFandG.
Similarly, ifzis a fixed point ofG, then it is also a fixed point ofF. Now letx0be an arbitrary point ofX. Ifx0 Fx0, the proof is finished, so assumex0/Fx0. We can define a sequence {xn}inXas follows:
x2n1Fx2n, x2n2Gx2n1 forn∈ {0,1, . . .}. 4.12 Without loss of generality, we can suppose that the successive terms of{xn} are different.
Otherwise, we are again finished. Note that sinceFandGare weakly increasing, we have x1Fx0GFx0 Gx1x2,
x2Gx1FGx1Fx2x3 4.13
and continuing this process, we have
x1x2 · · · xnxn1 · · ·. 4.14 Now sincex2n−1andx2nare comparable then, we can use the inequality4.7for these points then we have
TdFx2n, Gx2n−1, dx2n, x2n−1, dx2n, Fx2n, dx2n−1, Gx2n−1,
dx2n, Gx2n−1, dx2n−1, Fx2n≤0 4.15
and so
Tdx2n1, x2n, dx2n, x2n−1, dx2n, x2n1, dx2n−1, x2n,0, dx2n−1, x2n1≤0. 4.16
Now usingT1, we have
Tdx2n1, x2n, dx2n, x2n−1, dx2n, x2n1, dx2n−1, x2n,0, dx2n−1, x2n dx2n, x2n1≤0, 4.17 and formT2there exists a right continuous functionf :R → R, f0 0, ft< t,fort >0, we have for alln∈ {1,2, . . .}
dx2n1, x2n≤fdx2n, x2n−1. 4.18
Similarly, sincex2n andx2n1 are comparable, then we can use the inequality4.7for these points then we have
TdFx2n, Gx2n1, dx2n, x2n1, dx2n, Fx2n, dx2n1, Gx2n1,
dx2n, Gx2n1, dx2n1, Fx2n≤0 4.19
and so
Tdx2n1, x2n2, dx2n, x2n1, dx2n, x2n1, dx2n1, x2n2, dx2n, x2n2,0≤0. 4.20
Now again usingT1, we have
Tdx2n1, x2n2, dx2n, x2n1, dx2n, x2n1, dx2n1, x2n2,
dx2n, x2n1 dx2n1, x2n2,0≤0, 4.21
and formT2, we have for alln∈ {1,2, . . .},
dx2n1, x2n2≤fdx2n, x2n1. 4.22
Therefore, from4.18and4.22, we can have, for alln∈ {2,3, . . .}
dxn1, xn≤fdxn, xn−1 4.23
and so
dxn1, xn≤fn−1dx2, x1. 4.24
Thus fromLemma 3.1, we have, sincedx2, x1>0,
nlim→ ∞dxn1, xn 0. 4.25
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
d
x2nk, x2mk
≥δ fork∈ {1,2, . . .}. 4.26
We may also assumethat
d
x2mk−2, x2nk
< δ 4.27
by choosing 2mkto be the smallest number exceeding 2nkfor which4.26holds. Now 4.24,4.26, and4.27imply
0< δ≤d
x2nk, x2mk
≤d
x2nk, x2mk−2 d
x2mk−2, x2mk−1 d
x2mk−1, x2mk
≤δf2mk−3dx2, x1 f2mk−2dx2, x1
4.28
and so
klim→ ∞d
x2nk, x2mk
δ. 4.29
Also, by the triangular inequality, d
x2nk, x2mk−1
−d
x2nk, x2mk≤d
x2mk−1, x2mk
≤f2mk−2dx2, x1, d
x2nk1, x2mk−1
−d
x2nk, x2mk≤d
x2mk−1, x2mk d
x2nk, x2nk1
≤f2mk−2dx2, x1 f2nk−1dx2, x1.
4.30
Therefore, we get
klim→ ∞d
x2nk, x2mk−1 δ,
k→ ∞limd
x2nk1, x2mk−1
δ. 4.31
Also we have
δ≤d
x2nk, x2mk
≤d
x2nk, x2nk1 d
x2nk1, x2mk
≤f2nk−2dx2, x1 d
Fx2nk, Gx2mk−1 .
4.32
On the other hand, sincex2nk andx2mk−1are comparable, we can use the condition4.7 for these points. Therefore, we have
T d
Fx2nk, Gx2mk−1 , d
x2nk, x2mk−1 , d
x2nk, Fx2nk , d
x2mk−1, Gx2mk−1 , d
x2nk, Gx2mk−1 , d
x2mk−1, Fx2nk
≤0 4.33
and so
T d
x2nk1, x2mk , d
x2nk, x2mk−1 , d
x2nk, x2nk1 , d
x2mk−1, x2mk , d
x2nk, x2mk , d
x2mk−1, x2nk1
≤0. 4.34
Now, considering 4.29and 4.31and lettingk → ∞in the last inequality, we have, by continuity ofT, that
T
klim→ ∞d
x2nk1, x2mk
, δ,0,0, δ, δ ≤0. 4.35
FromT2, we have limk→ ∞dx2nk1, x2mk ≤ fδ. Therefore, lettingk → ∞in4.32, we haveδ≤fδ. This is a contradiction sinceft< tfort >0. Thus{x2n}is a Cauchy sequence inX, so{xn}is a Cauchy sequence. Therefore, there exists anx∈Xwith limn→ ∞xnx.
If4.8or4.9hold then clearlyx Fx Gx. Now suppose4.10holds. Suppose dx, Fx > 0. Now since limn→ ∞xn x, then from 4.10, x2n−1 x for all n. Using the inequality4.7, we have
TdFx, Gx2n−1, dx, x2n−1, dx, Fx, dx2n−1, Gx2n−1, dx, Gx2n−1, dx2n−1, Fx≤0.
4.36
So lettingn → ∞from the last inequality, we have
TdFx, x,0, dx, Fx,0,0, dx, Fx≤0 4.37
which is a contradiction toT3. Thusdx, Fx 0 and soxFxGx.
Remark 4.7. We can have some new results fromTheorem 4.5with some examples forT.
For example, we can have the following corollary.
Corollary 4.8. LetX,be a partially ordered set and suppose that there is a metricdonX such thatX, dis a complete metric space. SupposeF, G:X → Xare two weakly increasing mappings such that for all comparablex, y∈X,
d
Fx, Gy
≤φ d
x, y
, 4.38
whereφ:R → Ris a right continuous function such thatφ0 0,φt< tfort >0. Also
F or Gis continuous 4.39
or
if {xn} ⊂X is a nondecreasing sequence withxn−→xinX,
thenxnx ∀n 4.40
hold, thenFandGhave a common fixed point.
Proof. Let Tt1, . . . , t6 t1 −φt2, then it is obvious that T ∈ T. Therefore, the proof is complete fromTheorem 4.5.
5. Application
Consider the integral equations
xt b
a
K1t, s, xsdsgt, t∈a, b, xt
b
a
K2t, s, xsdsgt, t∈a, b.
5.1
The purpose of this section is to give an existence theorem for common solution of 5.1usingCorollary 4.8. This section is related to those19–22.
Letbe a partial order relation onRn. Theorem 5.1. Consider the integral equations5.1.
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
;
5.2
iiithere exist a continuous function p : a, b×a, b → R and a right continuous and nondecreasing functionφ:R → Rsuch thatφ0 0 andφt< tfort >0, such that
|K1t, s, u−K2t, s, v| ≤pt, sφ|u−v| 5.3
for eacht, s∈a, band comparableu, v∈Rn; ivsupt∈a,bb
apt, sds≤1.
Then the integral equations5.1have a unique common solutionx∗inCa, b,Rn.
Proof. LetX :Ca, b,Rnwith the usual supremum norm, that is,xmaxt∈a,b|xt|, for x∈Ca, b,Rn. Consider onXthe partial order defined by
x, y∈Ca, b,Rn, xy iffxtyt for any t∈a, b. 5.4
ThenX,is a partially ordered set. AlsoX, · 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 toxandy6.
DefineF, G:X → X, by Fxt
b
a
K1t, s, xsdsgt, t∈a, b, Gxt
b
a
K2t, s, xsdsgt, t∈a, b.
5.5
Now fromii, we have, for allt∈a, b, Fxt
b
a
K1t, s, xsdsgt
b
a
K2
t, s,
b
a
K1s, τ, xτdτgs
dsgt
b
a
K2t, s, Fxsdsgt
GFxt,
Gxt b
a
K2t, s, xsdsgt
b
a
K1
t, s,
b
a
K2s, τ, xτdτgs
dsgt
b
a
K1t, s, Gxsdsgt
FGxt.
5.6
Thus, we haveFxGFxandGx FGxfor allx∈X. This shows thatFandGare weakly increasing. Also for each comparablex, y∈X, we have
Fxt−Gyt
b
a
K1t, s, xsds− b
a
K2
t, s, ys ds
≤ b
a
K1t, s, xs−K2
t, s, ysds
≤ b
a
pt, sφxs−ysds
≤φx−yb
a
pt, sds
≤φx−y, for anyt∈a, b.
5.7
HenceFx−Gy ≤ φx−yfor each comparablex, y ∈ X. Therefore, all conditions of Corollary 4.8are satisfied. Thus the conclusion follows.
Acknowledgment
The authors thank the referees for their appreciation, valuable comments, and suggestions.
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