# Contractive Mapping in Generalized, Ordered Metric Spaces with Application in Integral Equations

Authors: L. Gholizadeh; R. Saadati; W. Shatanawi; S. M. Vaezpour
Journal: Mathematical Problems in Engineering
 (2011)

---

## Abstract

We consider the concept ofΩ-distance on a complete, partially ordered G-metric space and prove some fixed point theorems. Then, we present some applications in integral equations of our obtained results.


---

## Body

## 1. Introduction

The Banach fixed point theorem for contraction mapping has been generalized and extended in many directions [1–11]. Nieto and Rodríguez-López [10], Ran and Reurings [12], and Petrusel and Rus [13] presented some new results for contractions in partially ordered metric spaces. The main idea in [10, 12, 14] involves combining the ideas of an iterative technique in the contraction mapping principle with those in the monotone technique. Also, Mustafa and Sims [15] introduced the concept of G-metric. Some authors [16, 17] have proved some fixed point theorems in these spaces. Recently, Saadati et al. [18], using the concept of G-metric, defined an Ω-distance on complete G-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 see [15–18, 20–23].Definition 1 (see [15]).
LetX be a nonempty set. A function G:X×X×X→[0,∞) is called a G-metric if the following conditions are satisfied:(i)
G(x,y,z)=0 if x=y=z (coincidence),(ii)
G(x,x,y)>0 for all x,y∈X, where x≠y,(iii)
G(x,x,z)≤G(x,y,z) for all x,y,z∈X, with z≠y,(iv)
G(x,y,z)=G(p{x,y,z}), where p is a permutation of x,y,z (symmetry),(v)
G(x,y,z)≤G(x,a,a)+G(a,y,z) for all x,y,z,a∈X (rectangle inequality).A G-metric is said to be symmetric if G(x,y,y)=G(y,x,x) for all x,y∈X.Definition 2.
Let(X,G) be a G-metric space,(1)
a sequence{xn} in X is said to be G-Cauchy sequence if, for each ɛ>0, there exists a positive integer n0 such that for all m,n,l≥n0,   G(xn,xm,xl)<ɛ;(2)
a sequence{xn} in X is said to be G-convergent to a point x∈X if, for each ɛ>0, there exists a positive integer n0 such that for all m,n≥n0, G(xm,xn,x)<ɛ.Definition 3 (see [15]).
Let(X,G) be a G-metric space. Then a function Ω:X×X×X→[0,∞) is called an Ω-distance on X if the following conditions are satisfied: (a)
Ω(x,y,z)≤Ω(x,a,a)+Ω(a,y,z) for all x,y,z,a∈X,(b)
for anyx,y∈X,Ω(x,y,·),Ω(x,·,y):X→[0,∞) are lower semicontinuous,(c)
for eachɛ>0, there exists a δ>0 such that Ω(x,a,a)≤δ and Ω(a,y,z)≤δ imply G(x,y,z)≤ɛ.Example 1 (see [18]).
Let(X,d) be a metric space and G:X3→[0,∞) defined by
(1.1)G(x,y,z)=max⁡{d(x,y),d(y,z),d(x,z)},
for all x,y,z∈X. Then Ω=G is an Ω-distance on X.Example 2 (see [18]).
InX=ℝ we consider the G-metric G defined by
(1.2)G(x,y,z)=13(|x-y|+|y-z|+|x-z|),
for all x,y,z∈ℝ. Then Ω:ℝ3→[0,∞) defined by
(1.3)Ω(x,y,z)=13(|z-x|+|x-y|),
for all x,y,z∈ℝ is an Ω-distance on ℝ.For more example see [18].Lemma 1.1 (see [18]).
LetX be a metric space with metric G and Ω be an Ω-distance on X. Let xn,yn be sequences in X,αn,βn be sequences in [0,∞) converging to zero and let x,y,z,a∈X. Then one has the following.(1)
IfΩ(y,xn,xn)≤αn and Ω(xn,y,z)≤βn for n∈ℕ, then G(y,y,z)<ɛ and hence y=z.(2)
IfΩ(yn,xn,xn)≤αn and Ω(xn,ym,z)≤βn for m>n then G(yn,ym,z)→0 and hence yn→z.(3)
IfΩ(xn,xm,xl)≤αn for any l,m,n∈ℕ with n≤m≤l, then xn is a G-Cauchy sequence.(4)
IfΩ(xn,a,a)≤αn for any n∈ℕ then xn is a G-Cauchy sequence.Definition 4 (see [18]).
G-metric space X is said to be Ω-bounded if there is a constant M>0 such that Ω(x,y,z)≤M for all x,y,z∈X.


## 2. Fixed Point Theorems on Partially OrderedG-Metric Spaces

Definition 5.
Suppose(X,≤) is a partially ordered space and T:X→X is a mapping of X into itself. We say that T is nondecreasing if for x,y∈X,
(2.1)x≤y⟹T(x)≤T(y).Theorem 2.1.
Let(X,≤) be a partially ordered space. Suppose that there exists a G-metric on X such that (X,G) is a complete G-metric space and Ω is an Ω-distance on X such that X is Ω-bounded. Let f:X→X and g:X→X weakly compatible and f,g be non-decreasing mapping such that (a)
g(X)⊆f(X);(b)
Ω(gx,gy,gz)≤kmax{Ω(fx,fy,fz), Ω(fx,gx,fz), Ω(fy,gy,fz),Ω(fx,gy,fz),Ω(fy,gx,fz)};for all x,y,z∈Xand0≤k<1,(c)
for everyx∈X and y∈X with f(y)≠g(y), inf{Ω(fx,y,fx)+Ω(fx,y,gx)+Ω(fx,gx,y):f(x)≤g(x)}>0;(d)
there existx0∈X that f(x0)≤g(x0); then f and g have a unique common fixed point u in X and Ω(u,u,u)=0.Proof.
Letx0∈X that f(x0)≤g(x0). By part (a), we can choose x1∈X such that f(x1)=g(x0). Again from part (a), we can choose x2∈X such that f(x2)=g(x1). Continuing this process we can construct sequences {xn} in X such that,
(2.2)yn=gxn=fxn+1,∀n≥0,xn≤xn+1.
Now, since g is non-decreasing mapping then,
(2.3)gxn≤gxn+1,∀n≥0,
so, for all s≥0,
(2.4)Ω(yn,yn+1,yn+s)=Ω(gxn,gxn+1,gxn+s)≤kmax⁡{Ω(fxn,fxn+1,fxn+s),Ω(fxn,gxn,fxn+s),Ω(fxn+1,gxn+1,fxn+s),Ω(fxn,gxn+1,fxn+s),Ω(fxn+1,gxn,fxn+s)}=kmax⁡{Ω(yn-1,yn,yn+s-1),Ω(yn-1,yn,yn+s-1),Ω(yn,yn+1,yn+s-1),Ω(yn-1,yn+1,yn+s-1),Ω(yn,yn,yn+s-1)}.
Then,
(2.5)Ω(yn,yn+1,yn+s)≤kmax⁡{Ω(yn-1,yn,yn+s-1),Ω(yn,yn+1,yn+s-1),Ω(yn-1,yn+1,yn+s-1),Ω(yn,yn,yn+s-1)}.
Now since,
(2.6)Ω(yn-1,yn+1,yn+s-1)≤kmax⁡{Ω(yn-2,yn,yn+s-2),Ω(yn-2,yn-1,yn+s-2),Ω(yn,yn+1,yn+s-2),Ω(yn-2,yn+1,yn+s-2),Ω(yn,yn-1,yn+s-2)}Ω(yn,yn,yn+s-1)≤kmax⁡{Ω(yn-1,yn-1,yn+s-2),Ω(yn-1,yn,yn+s-2),Ω(yn-1,yn,yn+s-2),Ω(yn-1,yn,yn+s-2),Ω(yn-1,yn,yn+s-2)},
thus,
(2.7)Ω(yn,yn+1,yn+s)≤k2max⁡{Ω(yi,yj,yt),n-2≤i≤n,n-1≤j≤n+1,n+s-2≤t≤n+s-1}⋮≤kn-1max⁡{Ω(yi,yj,yt);1≤i≤n,2≤j≤n+1,s+1≤t≤n+s-1}.
So Ω(yn,yn+1,yn+s)≤kn-1Mn,s where
(2.8)Mn,s:=max⁡{Ω(yi,yj,yt),1≤i≤n,2≤j≤n+1,s+1≤t≤n+s-1}.
Now, for any l>m>n with m=n+k and l=m+t(k,t∈ℕ), we have,
(2.9)lim⁡m,n,l→∞Ω(yn,ym,yl)=0.
Since X is Ω-bounded and
(2.10)Ω(yn,ym,yl)≤Ω(yn,yn+1,yn+1)+Ω(yn+1,ym,yl)≤Ω(yn,yn+1,yn+1)+Ω(yn+1,yn+2,yn+2)+⋯+Ω(ym-1,ym,yl)≤kn-1Mn,1+knMn+1,2+⋯+km-2Mm-1,t+1≤∑j=1n-m+2kn-jM≤kn-11-kM,
so, by Part (3) of Lemma 1.1, {yn} is a G-Cauchy sequence. Since X is G-complete, {yn} converges to a point y∈X. Thus, for ɛ>0 and by the lower semicontinuity of Ω, we have
(2.11)Ω(yn,ym,y)≤lim⁡inf⁡p→∞Ω(yn,ym,yp)≤ɛ,m≥nΩ(yn,y,yl)≤lim⁡inf⁡p→∞Ω(yn,yp,yl)≤ɛ,l≥n.
Assume that fy≠gy. Since,
(2.12)yn=fxn+1=gxn≤gxn+1=fxn+2=yn+1,
so, yn≤yn+1, and,
(2.13)0<inf⁡{Ω(yn,y,yn)+Ω(yn,yn+1,y)+Ω(yn,y,yn+1)}≤3ɛ,
for every ɛ>0, that is a contraction. So, we have fy=gy. Then, by (b),
(2.14)Ω(gy,gy,gy)≤kΩ(gy,gy,gy),
so, Ω(gy,gy,gy)=0. Similarly, Ω(g2y,g2y,gy)=0.
Now,(2.15)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)}.
Thus,
(2.16)Ω(gy,g2y,gy)=0,Ω(g2y,gy,gy)=0.
By Part (c) of Definition 3, G(g2y,g2y,gy)=0 and consequently g2y=gy which implies that gy is a fixed point for g. Now,
(2.17)f(gy)=g(fy)=g2y=gy.
So, it is enough to put gy=u, then u is a common fixed point of f and g.
Uniqueness: Assume that there exist v∈X such that fv=gv=v. Hence, we have,
(2.18)Ω(v,v,v)≤kΩ(v,v,v),
and so Ω(v,v,v)=0. Also, Ω(v,v,u)=0. On the other hand,
(2.19)Ω(v,u,u)≤kmax⁡{Ω(v,u,u),Ω(u,v,u)},Ω(u,v,u)≤kmax⁡{Ω(u,v,u),Ω(v,u,u)},
which follows that, Ω(v,u,u)=Ω(u,v,u)=0. Then by Part (c) of Definition 3, u=v and Ω(u,u,u)=0.The following corollary is a generalization of [24, Theorem 2.1].Corollary 2.2.
Let(X,≤) be a partially ordered space. Suppose that there exists a G-metric on X such that (X,G) is a G-metric space and Ω is an Ω-distance on X such that X be Ω-bounded. Let f:X→X and g:X→X be weakly compatible and f,g be a non-decreasing mapping such that (a)
g(X)⊆f(X) and either f(X) or g(X) is complete;(b)
for allx,y,z∈X and 0≤k<1, Ω(gx,gy,gz)≤kΩ(fx,fy,fz);(c)
for everyx∈X and y∈X with f(y)≠g(y),  inf{Ω(fx,y,fx)+Ω(fx,y,gx)+Ω(fx,gx,y):f(x)≤g(x)}>0;(d)
there existx0∈X that f(x0)≤g(x0);then f and g have a unique common fixed point y in X and Ω(y,y,y)=0.Definition 6 (see [25]).
LetΦ be the set of all functions φ such that φ:[0,∞)→[0,∞) is a continuous and nondecreasing function with φ(t)<tfor allt∈ℝ+ and ∑n=1∞φn(t)<∞for eacht∈ℝ+. The function φ is called a growth or control function of T:X→X.
It is clear that(2.20)lim⁡n→∞φn(t)=0,∀t∈R+,φn(0)=0.Theorem 2.3.
Let(X,≤) be a partially ordered space. Suppose that there exists a G-metric on X such that (X,G) is a complete G-metric space and Ω is an Ω-distance on X and T is a non-decreasing mapping from X into itself. Let X be Ω-bounded. Suppose that φ∈Φ and
(2.21)Ω(Tx,T2x,Tw)≤φ(Ω(x,Tx,w))∀x≤Tx,w∈X.
Also, for every x∈X(2.22)inf⁡{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,T2x,y):x≤Tx}>0,
for every y∈X with y≠Ty. If there exists an x0∈X with x0≤Tx0, then T has a unique fixed point. Moreover, if v=Tv, then Ω(v,v,v)=0.Proof.
Ifx0=Tx0, then the proof is finished. Suppose that Tx0≠x0. since x0≤Tx0 and T is non-decreasing, we obtain
(2.23)x0≤Tx0≤T2x0≤⋯≤Tn+1x0≤⋯
For all n∈ℕ and t≥0,
(2.24)Ω(Tnx0,Tn+1x0,Tn+tx0)≤φ(Ω(Tn-1x0,Tnx0,Tn+t-1x0))≤φ2(Ω(Tn-2x0,Tn-1x0,Tn+t-2x0))⋮≤φn(Ω(x0,Tx0,Ttx0)).
We claim that for m=n+k and l=m+t(k,t∈ℕ) with l>m>n,
(2.25)lim⁡m,n,l→∞Ω(Tnx0,Tmx0,Tlx0)=0.
We prove by,
(2.26)Ω(Tnx0,Tmx0,Tlx0)≤Ω(Tnx0,Tn+1x0,Tn+1x0)+Ω(Tn+1x0,Tmx0,Tlx0)≤Ω(Tnx0,Tn+1x0,Tn+1x0)+Ω(Tn+1x0,Tn+2x0,Tn+2x0)+⋯+Ω(Tm-1x0,Tmx0,Tlx0)≤φn(Ω(x0,Tx0,Tx0))+φn+1(Ω(x0,Tx0,Tx0))+⋯+φm-2(Ω(x0,Tx0,Tx0))+φm-1(Ω(x0,Tx0,Tt+1x0))≤φn-1(M)(∑n=1∞φn(M)).
Since ∑n=1∞φn(M)<∞, so,
(2.27)lim⁡m,n→∞Ω(Tnx0,Tmx0,Tlx0)=0.
By Part (c) of Lemma 1.1{Tnx0} is a G-Cauchy sequence. Since X is G-complete, {Tnx0} converges to a point u∈X. Let n∈ℕ be fixed. By lower semicontinuity of Ω,
(2.28)Ω(Tnx0,Tmx0,u)≤lim⁡inf⁡p→∞Ω(Tnx0,Tmx0,Tpx0)≤ɛ,m>n,Ω(Tnx0,u,Tlx0)≤lim⁡inf⁡p→∞Ω(Tnx0,Tpx0,Tmx0)≤ɛ,l≥n.
Assume that u≠Tu. Since Tnx0≤Tn+1x0,
(2.29)0<inf⁡{Ω(Tnx0,u,Tnx0)+Ω(Tnx0,u,Tn+1x0)+Ω(Tnx0,Tn+2x0,u):n∈N}≤3ɛ,
for every ɛ>0, which is a contraction. Therefore, we have u=Tu.
Uniqueness: let v be another fixed point of T, then
(2.30)Ω(u,u,v)=Ω(Tu,T2u,Tv)≤φ(Ω(u,Tu,v))<Ω(u,u,v),
which is a contraction. Therefore, fixed point u is unique. Now, if v=Tv, we have,
(2.31)Ω(v,v,v)=Ω(Tv,T2v,T3v)≤φ(Ω(v,Tv,T2v))=φ(Ω(v,v,v)).
So Ω(v,v,v)=0.Corollary 2.4.
Let the assumptions of Theorem2.3 hold and
(2.32)Ω(Tmx,Tm+1x,Tmw)≤φ(Ω(x,Tx,w))∀m∈N,x≤Tx,w∈X,
then T has a unique fixed point.Proof.
From Theorem2.3, Tm has a unique fixed point u. However,
(2.33)Tu=T(Tmu)=Tm+1u=TmTu,
so Tu is also a fixed point of Tm. Since the fixed point of Tm is unique, it must be the case that Tu=u.Corollary 2.5.
Let the assumptions of Theorem2.3 hold and T:X→X satisfies,
(2.34)Ω(Tx,T2x,Tx)≤φ(Ω(x,Tx,x))∀x≤Tx.
Then T has a unique fixed point.Proof.
Takew=x, and apply Theorem 2.3.Theorem 2.6.
Let(X,≤) be a partially ordered space. Suppose that there exists a G-metric on X such that (X,G) is a complete G-metric space, Ω is an Ω-distance on X, and T is a non-decreasing mapping from X into itself. Let X be Ω-bounded. Suppose that
(2.35)Ω(Tx,T2x,Tw)≤k(Ω(x,T2x,Tw)+Ω(x,Tx,Tx)),
where x≤Tx,w∈X,k∈[0,1/3). Also for every x∈X,
(2.36)inf⁡{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,T2x,y):x≤Tx}>0,
for every y∈X with y≠Ty. If there exists an x0∈X with x0≤Tx0, then T has a unique fixed point say u and Ω(u,u,u)=0.Proof.
Letx0∈X be an arbitrary point, and define the sequence xn by xn=Tnx0. By (2.35) and for all t≥0,
(2.37)Ω(xn,xn+1,xn+t)≤k(Ω(xn-1,xn+1,xn+t)+Ω(xn-1,xn,xn)).
But by Part (a) of Definition 3,
(2.38)Ω(xn-1,xn+1,xn+t)≤Ω(xn-1,xn,xn)+Ω(xn,xn+1,xn+t).
Hence,
(2.39)Ω(xn,xn+1,xn+t)≤k[2Ω(xn-1,xn,xn)+Ω(xn,xn+1,xn+t)],
which implies,
(2.40)Ω(xn,xn+1,xn+t)≤2k1-kΩ(xn-1,xn,xn).
Let r=2k/(1-k), then r<1 and by repeated application of (2.40), we have
(2.41)Ω(xn,xn+1,xn+t)≤rnΩ(x0,x1,x1).
Now, for any l>m>n with m=n+k and l=m+t(k,t∈ℕ), we have,
(2.42)Ω(xn,xm,xl)≤Ω(xn,xn+1,xn+1)+Ω(xn+1,xn+2,xn+2)+Ω(xn+2,xn+3,xn+3)+⋯+Ω(xm-1,xm,xl)≤(rn+rn+1+⋯+rm-1)Ω(x0,x1,x1)≤rn1-rΩ(x0,x1,x1).
So,
(2.43)lim⁡m,n,l→∞Ω(xn,xm,xl)=0.
By Part (3) of Lemma 1.1, xn is a G-Cauchy sequence. Since X is G-complete, xn converges to a point u∈X. Now, similar to proving Theorem 2.1, T has a unique fixed point and Ω(u,u,u)=0.Corollary 2.7.
Let the assumptions of Theorem2.6 hold and
(2.44)Ω(Tmx,Tm+2x,Tmw)≤k(Ω(x,Tm+2x,Tmw)+Ω(x,Tmx,Tmx))
where k∈[0,1/3), then T has a unique fixed point.Proof.
The argument is similar to that used in the proof of Corollary2.4.


## 3. Applications

In this section, we give an existence theorem for a solution of a class of integral equations. Denote byΛ the set of all functions λ:[0,+∞)→[0,+∞) satisfying the following hypotheses: (i)
λ is a Lebesgue-integrable mapping on each compact of [0,+∞),(ii)
for everyϵ>0, we have ∫0ϵλ(s)ds>0,(iii)
∥λ∥<1, where ∥λ∥ denotes to the norm of λ.Now, we have the following results.Theorem 3.1.
Let(X,≤) be a partially ordered space. Suppose that there exists a G-metric on X such that (X,G) is a complete G-metric space and Ω is an Ω-distance on X and T is a non-decreasing mapping from X into itself. Let X be Ω-bounded. Suppose that
(3.1)Ω(Tx,T2x,Tw)≤∫0Ω(x,Tx,w)α(s)ds,
where α∈Λ. Also, suppose that for every x∈X(3.2)inf⁡{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,T2x,y):x≤Tx}>0,
for every y∈X with y≠Ty. If there exists an x0∈X with x0≤Tx0, then T has a unique fixed point.Proof.
Defineϕ:[0,+∞)→[0,+∞) by ϕ(t)=∫0tα(s)ds. It is clear that ϕ is nondecreasing and continuous. From (iii), we have
(3.3)ϕ(t)=|ϕ(t)|=|∫0tλ(s)ds|≤∫0t|λ(s)|ds≤‖λ‖t<t.
Also, note that
(3.4)ϕ2(t)=ϕ(ϕ(t))≤‖λ‖ϕ(t)≤‖λ‖2t.
In general, we have ϕn(t)≤∥λ∥nt. Thus, we have
(3.5)∑n=1∞ϕn(t)≤∑n=1∞‖λ‖nt=‖λ‖t1-‖λ‖<+∞.
Therefore ϕ satisfies all the hypotheses of Definition 6. By inequality (3.1), we have Ω(Tx,T2x,Tw)≤ϕ(Ω(x,Tx,w). Therefore by Theorem 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:(3.6)u(t)=∫01K(t,s,u(s))ds+g(t),t∈[0,1].
Let X=C([0,1]) be the set of all continuous functions defined on [0,1]. Define (3.7)G:X×X×X⟶R+
by (3.8)G(x,y,z)=max⁡{‖x-y‖,‖x-z‖,‖y-z‖},
where ∥x∥=sup{|x(t)|:t∈[0,1]}. Then (X,G) is a complete G-metric space. Let Ω=G. Then Ω is an Ω-distance on X.Define an ordered relation ≤ onX by (3.9)x≤yiffx(t)≤y(t),∀t∈[0,1].
Then (X,≤) is a partially ordered set. Now, we prove the following result.Theorem 3.2.
Suppose the following hypotheses hold.(a)
K:[0,1]×[0,1]×ℝ+→ℝ+ and g:ℝ+→ℝ+ are continuous.(b)
K is nondecreasing in its first coordinate and g is nondecreasing.(c)
There exist a continuous functionG:[0,1]×[0,1]→[0,+∞] such that
(3.10)|K(t,s,u)-K(t,s,v)|≤G(t,s)|u-v|,for each comparableu,v∈ℝ+ and each t,s∈[0,1].(d)
supt∈[0,1]∫01G(t,s)ds≤r for some r<1. Then the integral equation (3.6) has a solution u∈C([0,1]).Proof.
DefineT:C([0,1])→C([0,1]) by
(3.11)Tx(t)=∫01K(t,s,x(s))ds+g(t),t∈[0,1].
By hypothesis (b), we have that T is nondecreasing.
Now, if(3.12)inf⁡{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,T2x,y):x≤Tx}=0,
for y∈C([0,1]) with y≠Ty, then for each n∈ℕ there exists xn∈C([0,1]) with xn≤Txn such that
(3.13)Ω(xn,y,xn)+Ω(xn,y,Txn)+Ω(xn,T2xn,y)≤1n.
So, we have
(3.14)Ω(xn,y,Txn)=max⁡{‖xn-y‖,‖xn-Txn‖,‖y-Txn‖}≤1n.
Therefore, for each t∈[0,1], we have
(3.15)lim⁡n→+∞xn(t)=y(t),lim⁡n→+∞Txn(t)=y(t).
By the continuity of K, we have
(3.16)y(t)=lim⁡n→+∞Txn(t)=∫01K(t,s,lim⁡n→+∞xn(s))ds+g(t)=∫01K(t,s,y(s))ds+g(t)=Ty(t).
Thus, we have y=Ty, a contradiction. Thus,
(3.17)inf⁡{Ω(x,y,x)+Ω(x,y,Tx)+Ω(x,T2x,y):x≤Tx}>0.
Define ϕ:[0,+∞)→[0,+∞) by ϕ(t)=rt. For x∈C([0,T]) with x≤Tx, we have
(3.18)Ω(Tx,T2x,Tx)=sup⁡t∈[0,1]|Tx(t)-T2x(t)|=sup⁡t∈[0,1]|∫01K(t,s,x(s))-K(t,s,Tx(s))ds|≤sup⁡t∈[0,1]∫01|K(t,s,x(s))-K(t,s,Tx(s))|ds≤sup⁡t∈[0,1]∫01G(t,s)|x(s)-Tx(s)|ds≤sup⁡t∈[0,1]|x(t)-Tx(t)|sup⁡t∈[0,1]∫0TG(t,s)ds=Ω(x,Tx,x)sup⁡t∈[0,1]∫01G(t,s)ds≤rΩ(x,Tx,x)=ϕ(Ω(x,Tx,x)).
Moreover, take x0=0, then x0≤Tx0. Thus all the required hypotheses of Corollary 2.5 are satisfied. Thus there exists a solution u∈C([0,T]) of the integral equation (3.6).



---

*Source: 380784-2011-11-17.xml*