# Fixed Points and Continuity for a Pair of Contractive Maps with Application to Nonlinear Volterra Integral Equations

Authors: Santosh Kumar
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we have established and proved fixed point theorems for the Boyd-Wong-type contraction in metric spaces. In particular, we have generalized the existing results for a pair of mappings that possess a fixed point but not continuous at the fixed point. We can apply this result for both continuous and discontinuous mappings. We have concluded our results by providing an illustrative example for each case and an application to the existence and uniqueness of a solution of nonlinear Volterra integral equations.


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## Body

## 1. Introduction and Preliminaries

Continuity is an ideal property which is sometimes difficult to be fulfilled especially in some daily life applications. For instance, most neural network systems like bar code scanning, speech recognition, and handwritten digit recognition lack the continuity property. These neural network systems are some excellent prototypes for learning discontinuity phenomena. Here, we transform different kinds of day to day real-world phenomena into threshold functions which satisfies our desirable continuity of the weaker form and a new type of contraction to provide a solution to some daily life applications. Therefore, it is desirable to relax continuity assumptions because, in some applications, the function may not be continuous. One can see more literature on the topic [1–5].In 1969, Kannan [6] proved the following fixed point theorem for discontinuous mapping:Theorem 1 [6].
If a self mappingT of a complete metric space X,d satisfies the condition
(1)dTx,Ty≤adx,Tx+dy,Ty,0≤a<12,for each x,y∈X, then, T has a unique fixed point.This theorem gave rise to the question of continuity of contractive mappings at their fixed points. In the Kannan contractive condition, continuity of mappingT was not required for the existence of a fixed point.In 1971, Ciric [7] (see also [8]) introduced the notion of orbital continuity, which is as follows:Definition 2 (see [7]).
LetX,d be a metric space and T:X⟶X be a self mapping. Then, the set Ox,T=Tnx:n=0,1,2,3,⋯, is called the orbit of T at x and T is called orbitally continuous if for any sequence xn⊂Ox,T, xn⟶z implies that Txn⟶Tz as n⟶∞.In 2017, Pant and Pant [9] introduced the notion of k-continuity which is as follows:Definition 3 [9].
A mappingT:X⟶X is called k-continuous for k=1,2,3,⋯, if Tkxn⟶Tt whenever a sequence xn is in X such that Tk−1xn⟶t.Continuity ofT implies orbital continuity, but the converse is not true (see [7]).The following are the examples ofk-continuity:Example 1.
LetX=0,3 be equipped with the usual metric and T:X⟶X be defined by
(2)Tx=1,if0≤x≤1,0,ifx∈1,3.
Then,Txn⟶t⇒T2xn⟶t, since Txn⟶t implies that t=0 or t=1 and T2xn=1 for all n, that is, T2xn⟶1=Tt. Hence, T is 2-continuous. However, T is discontinuous at x=1.Example 2.
LetX=0,5 be equipped with the usual metric and T:X⟶X be defined by
(3)Tx=1,if0≤x≤1,0,if1<x≤4,x4,if4<x≤5.
Then,T2xn⟶t⇒T3xn⟶Tt, since T2xn⟶t implies t=0 or t=1 and T3xn=1=Tt for each n. Hence, T is 3-continuous. However, Txn⟶t does not imply that T2xn⟶Tt, that is, T is not 2-continuous.Example 3 [9].
LetX=0,2 be equipped with the usual metric and T:X⟶X be defined by
(4)Tx=1+x2,if0≤x≤1,0,ifx>1.
Then, it can be verified thatT is 2-continuous but not continuous. Moreover, Tk is discontinuous for each positive integer k. Thus 2-continuity of T does not imply continuity of T2. In general, k-continuity of T does not imply continuity of Tn.Example 4 [9].
LetX=0,3∪4,5 be equipped with the usual metric and T:X⟶X be defined by
(5)Tx=1,if0≤x≤1,0,if1<x≤3,x4,if4<x<5.
Then,T2 is continuous but T is not 2-continuous. If we consider the sequence xn given by xn=4+1/n, then, Txn⟶1 but T2xn⟶0≠T1. Hence, T is not 2-continuous.From the above examples, one can see that continuity ofTk and k-continuity of T are independent conditions when k>1. It is easy to see that 1-continuity is equivalent to continuity and
(6)Continuity⇒2−continuity⇒3−continuity⇒⋯.Definition 4 [10].
Letxn be a sequence in a metric space X,d. Then,
(i)
A sequencexn in X converges to a point x∈X if and only if limn⟶∞dx,xn=limn⟶∞dxn,xn(ii)
A sequencexn is called a Cauchy sequence if there exists ε>0 such that for all n,m>N, we have dxn,xm<ε for some integers N≥0, that is limn,m⟶+∞dxn,xm exists and it is finite(iii)
A metric spaceX,d is complete if every Cauchy sequence xn converges to a point x∈X such that dx,x=limn,m⟶+∞dxn,xmPant and Pant [9] proved the following theorem by employing a new type of ε−δ condition.Theorem 5 [9].
Letf be a self mapping of a complete metric space X,d such that
(i)
dfx,fy<maxdx,fx,dy,fy, maxdx,fx,dy,fy>0(ii)
Given thatε>0, there exists a δ>0 such that ε<maxdx,fx,dy,fy≤ε+δ⇒dfx,fy≤ε
Iff is k-continuous or fk is continuous for some k≥1 or f is orbitally continuous, then, f possesses a unique fixed point.


## 2. Main Results

Pant and Pant [9] used ε−δ and k-continuity property to prove the above fixed point theorem for one self map. In this section, we are extending Theorem 5 for a pair of self maps using ε−δ conditions as follows:Theorem 6.
LetX be a nonempty set and let d to be a metric on X. Let T and S be self mappings of a complete metric space X,d satisfying
(i)
dTx,Sy<Mx,y, where
(7)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.(ii)
Given thatε>0, there exists a δ>0 such that ε<Mx,y≤ε+δ⇒dTx,Sy≤ε
IfT and S are k-continuous or Tk and Sk are continuous for some k≥1 or T and S are orbitally continuous, then, T and S have a unique common fixed point.Proof.
Letx0 be any point in X. Define a sequence xninXforn=0,1,2,⋯, as x2n+1=Tx2n and x2n+2=Sx2n+1, for all integers n≥0.
Ifx2n=x2n+1 for some n, then, x2n=Tx2n, that is x2n is a fixed point of T. Similarly, if there exists an integer N≥0 such that x2N+1=x2N+2, then, x2n+1 is a fixed point of S. This concludes the proof.☐Otherwise, we suppose thatx2n≠x2n+1, for all integers n≥0. Let d2n=dx2n,x2n+1; obviously, d2n+1=dx2n+1,x2n+2.Then, by using equation (7) with x=x2n and y=x2n+1, we have
(8)dx2n+1,x2n+2<Mx2n,x2n+1,where
(9)Mx2n,x2n+1=maxdx2n,x2n+1,dx2n,Tx2n,dx2n+1,Sx2n+1,dx2n,Sx2n+1+dx2n+1,Tx2n2,=maxdx2n,x2n+1,dx2n,x2n+1,dx2n+1,x2n+2,dx2n,x2n+2+dx2n+1,x2n+12.Since,(10)dx2n,x2n+2+dx2n+1,x2n+12≤dx2n,x2n+1+dx2n+1,x2n+2+dx2n+1,x2n+12=dx2n,x2n+1+dx2n+1,x2n+22,then,
(11)Mx2n,x2n+1=maxdx2n,x2n+1,dx2n+1,x2n+2.Thus,(12)dx2n+1,x2n+2=dTx2n,Sx2n+1≤maxdx2n,x2n+1,dx2n+1,x2n+2.Obviously, ifmaxdx2n,x2n+1,dx2n+1,x2n+2=dx2n+1,x2n+2, we have a contradiction and so maxdx2n,x2n+1,dx2n+1,x2n+2=dx2n,x2n+1. Therefore,
(13)dx2n+1,x2n+2<dx2n,x2n+1,which implies that the sequence d2n is decreasing to a nonnegative real number, say ε, for all integers n≥0. We claim that ε=0. In contrary, suppose that ε>0. Taking the limit as n⟶∞ in (13), we obtain
(14)ε<d2n≤ε+δ⟹d2n+1≤ε,which is a contradiction; hence, we conclude that ε=0 and
(15)limn⟶∞d2n=limn⟶∞dx2n,x2n+1=0.Now, we need to show that a sequencex2ninX is a Cauchy sequence. In equation (15), it is sufficient to show that a subsequence x2nr is a Cauchy sequence. On the contrary, we claim that x2nr is not a Cauchy sequence. Therefore, there exists ε>0 and a sequence of integers mr,nr such that
(16)dx2nr,x2mr≥ε,for all nr>mr≥r for some r≥0.Furthermore, suppose thatmr is the smallest integer which is chosen in such away that (16) holds so that we have
(17)dx2nr=dx2nr,x2mr−1<ε.Now, for allnr>mr, we have
(18)dx2nr,x2mr≤dx2nr,x2mr−1+dx2mr−1,x2mr.Asr⟶∞ in (18) and considering (15) and (17), we see that
(19)dx2nr,x2mr⟶ε.By similar computations, we see that,(20)dx2nr−1=dx2nr−1,x2mr−1⟶ε.To show it, we shall prove thatMx2nr−1,x2mr−1≤ε+δ. Then, by using equation (7) with x=x2nr−1andy=x2mr−1, we have
(21)dx2nr,x2mr=dTx2nr−1,Sx2mr−1≤Mx2nr−1,x2mr−1,where
(22)Mx2nr−1,x2mr−1=maxdx2nr−1,x2mr−1,dx2nr−1,Tx2nr−1,dx2mr−1,Sx2mr−1,dx2nr−1,Sx2mr−1+dx2mr−1,Tx2nr−12=maxdx2nr−1,x2mr−1,dx2nr−1,x2nr,dx2mr−1,x2mr,dx2nr−1,x2mr+dx2mr−1,x2nr2=dx2nr−1,x2mr−1.Asr⟶∞ in (22) and considering (19) and (20), then, (21) becomes
(23)ε<d2nr−1≤ε+δ⟹d2nr≤ε,which is a contradiction. Hence, x2ninX is a Cauchy sequence and
(24)limn,m⟶∞dx2n,x2m=0.SinceX is complete, there exists a point t∈X such that x2n⟶t. Furtheremore, for each k≥1, we have Tkx2n⟶Tt. Thus, t=Tt as Tkx2n⟶t. Hence, t is a fixed point of T.Again, forx2n+1⟶t and for each k≥1, we have Skx2n+1⟶St. Hence, t=St as Skx2n+1⟶t. Therefore, t is a fixed point of S.In addition, assume thatTk and Sk are k-continuous for some positive integer k. Then, we have limn⟶∞dx2n,t=0. Therefore,
(25)limn⟶∞Tkx2n=limn⟶∞Tx2n=limn⟶∞x2n+1=t,limn⟶∞Skx2n+1=limn⟶∞Sx2n+1=limn⟶∞x2n+2=t.Here, we will prove thatt is a fixed point of S. Contrarily, suppose that t≠St.Now,(26)dx2n+1,St=dTx2n,St≤Mx2n,t,where
(27)Mx2n,t=maxdx2n,t,dx2n,Tx2n,dt,St,dx2n,St+dt,Tx2n2=maxdx2n,t,dx2n,x2n+1,dt,St,dx2n,St+dt,x2n+12.Asn⟶∞ in (27), we see that
(28)Mx2n,t⟶dt,St.Applying the limit asn⟶∞ in (26), we have
(29)dt,St≤dt,St<dt,St,which is a contradiction. Hence, St=t.Now, suppose thatT is orbitally continuous. Since x2n⟶t, orbital continuity implies that Tx2n⟶Tt or Sx2n+1⟶St. This gives t=Tt as Tx2n⟶t or t=St as Sx2n+1⟶t. Thus, t is a fixed point of T and S.Next, we will show that a pointt is a unique common fixed of T and S. In contrary, suppose that t∈X and y∗∈X are two different common fixed points of T and S, respectively. Thus, dt,y∗>0.Now,(30)d=d≤M,where
(31)Mt,y∗=maxdt,y∗,dt,Tt,dy∗,Sy∗,dt,Sy∗+dy∗,Tt2=maxdt,y∗,dt,t,dy∗,y∗,dt,y∗+dy∗,t2=dx∗,y∗.Hence,(32)dt,y∗=dTt,Sy∗<Mdt,y∗<dt,y∗,which is a contradiction. Therefore, T and S have a unique common fixed point, that is t=y∗.To prove that any fixed point ofT is also a fixed point of S, conversely, we suppose to the contrary that t=Tt and t≠St. Now,
(33)dTt,St=dt,St≤Mt,St,where
(34)Mt,St=maxdt,St,dt,Tt,dSt,S2t,dt,S2t+dSt,Tt2=maxdt,St,dt,t,dSt,S2t,dt,S2t+dSt,t2=dt,St.Thus,(35)dTt,St=dt,St≤Mt,St<dt,St,which is a contradiction. Therefore, t=Tt=St. In a similar way, it is easy to show that any fixed point of S is also a fixed point of T.Remark 7.
If we setS=T, we get an improved version of the Bisht and Pant [11] theorem as a corollary for the case a=1 as follows:Corollary 8.
LetX be a nonempty set and let d to be a metric on X. Let T be self mapping of a complete metric space X,d satisfying
(i)
dTx,Ty<Mx,y, where
(36)Mx,y=maxdx,y,dx,Tx,dy,Ty,dx,Ty+dy,Tx2.(ii)
Given thatε>0, there exists a δ>0 such that ε<Mx,y≤ε+δ⇒dTx,Ty≤ε
IfT is k-continuous or Tk is continuous for some k≥1 or T is orbitally continuous, then, T has a fixed point.Corollary 9.
The conclusions of Theorem6 remain true, if we replace Mx,y in the contractive equation (7) by any one of the following:
(i)
Mx,y=maxdx,y,dx,Tx,dy,Sy.(ii)
Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.(iii)
Mx,y=maxdx,y,dx,Sx,dy,Ty,dx,Tx+dy,Sy2.
The following example shows the generality of Theorem6 over Theorem 5.Example 5.
LetX=0,3 be equipped with the usual metric. Let S and T be self mappings on X, i.e., T,S:X⟶X defined by
(37)Tx=1+x2,if0≤x≤1,0,ifx∈1,3.Sx=x2,if0≤x≤1,0,if1<x≤2,x2,if2<x≤3.Hence,T and S satisfy all the conditions of the above theorem and have a unique fixed point x=1; and S and T are discontinuous at x=1. The mapping T is 2-continuous and S is 3-continuous at x=1. S and T are orbitally continuous. Tt can be easily verified using the following cases:Case 1.
Now, we have(38)dTx,Sy<Mx,ywhere
(39)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.Forx,y≤1 and 0≤x≤1, we get
(40)dTx,Sy=d1+x2,y2=1+x2−y2=x−2y2+12,dx,y=dx,y=x−y,dx,Tx=dx,1+x2=x−1+x2=x−12,dy,Sy=dy,y2=y−y2,dx,Sy=dx,y2=x−y2,dy,Tx=dy,1+x2=y−1+x2=2y−x−12,Mx,y=maxx−y,x−12,y−y2,2x−y2+2y−x−14=x−y.By usingi of Theorem 6, we have
(41)dTx,Sy<Mx,y⟹x−2y2+12<x−y,which is a contradiction. Hence, T and S are discontinuous at x=1.Remark 10.
In case ofx=1 and y=1, both functions are discontinuous at this point but have the property of k-continuity. For more details, see Example 1.Case 2.
Next we have,(42)dTx,Sy<Mx,ywhere
(43)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.Forx,y>2⟹2<x≤3, we get
(44)dTx,Sy=d0,y2=0−y2=y2,dx,y=dx,y=x−y,dx,Tx=dx,0=x−0=x,dy,Sy=dy,y2=y−y2=y2,dx,Sy=dx,y2=x−y2=2x−y2,dy,Tx=dy,0=y−0=y,Mx,y=maxdx,y,dx,0,dy,y2,dx,y/2+dy,02,=maxx−y,x,y2,2x−y/2+y2=maxx−y,x,y2,2x+y4=x.By usingi of Theorem 6, we have
(45)dTx,Sy<Mx,y⟹y2<x.Thus, conditionsi of Theorem 6 satisfy for all x,y>2. This shows that T and S are continuous.Case 3.
Next, we have(46)dTx,Sy<Mx,ywhere
(47)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.Forx≤1⟹y≤2, we have
(48)dTx,Sy=d1+x2,0=1+x2−0=x+12,dx,y=dx,y=x−y,dx,Tx=dx,1+x2=x−1+x2=x−12,dy,Sy=dy,0=y−0=y,dx,Sy=dx,0=x−0=x,dy,Tx=dy,1+x2=y−1+x2=2y−x−12,Mx,y=maxdx,y,dx,1+x2,dy,0,dx,0+dy,1+x/22,=maxx−y,x−12,y,x+2y−x−1/22,=maxx−y,x−12,y,x+2y−14=y.By usingi of Theorem 6, we have
(49)dTx,Sy<Mx,y⟹x+12<y.Thus, conditionsi of Theorem 6 satisfy for all x≤1,y≤2. This shows that T and S are continuous. Also, T1=1 and S2=0. Then, TSx2n⟶t⟹T2Sx2n⟶1=TSt for each n. Hence, TS is 3-continuous.Therefore,T and S satisfy condition ii of Theorem 6 with δ=1−ε if ε<1 and δ=1 for ε≥1. To see this, consider Case 1, Case 2, and Case 3 as follows:By Case1, using condition ii of Theorem 6, ε<1 and δ=1−ε, we get
(50)ε<Mx,y<ε+δ⇒dTx,Sy<ε,⇒ε<x−12<ε+δ⇒x−2y2+12<ε,⇒ε<x−12<ε+1−ε⇒x−2y2+12<ε,⇒ε<0<1⇒0<ε.which is a contradiction.By Case2, using ii of Theorem 6, δ=1 for ε≥1, we get
(51)ε<Mx,y<ε+δ⇒dTx,Sy<ε,⇒ε<x<ε+δ⇒y2<ε,⇒ε<x<ε+1⇒y2<ε,satisfying for all x,y>2.However, this example is not applicable to the conditions imposed in Theorem5.Remark 11.
It can be seen from the above example thatT and S are threshold operation that models firing of a neuron, a function of two diodes, and also a low-pass filter that allows low voltages to pass but not higher voltages (e.g., noise in music systems).One of the fundamental tools for nonlinear analysis is the Banach fixed point theorem [12]. As a result of its usefulness and applications, this theorem has been massively investigated and generalized by different researchers. One of the important generalization of the Banach fixed point theorem is the Boyd and Wong [13] fixed point theorem. A mapping T satisfying
(52)dTx,Ty≤ϕdx,y,∀x,y∈M,whereby M,d is a complete metric space and a mapping ϕ:0,∞⟶0,∞ is upper semicontinuous from the right on 0,∞ such that ϕt<t, ∀t>0. Consequently, T has a unique fixed point z∈M and dTnx,z⟶0 as n⟶∞, ∀x∈M. Pant and Pant [9] proved the following theorem for the Boyd and Wong type fixed point theorem in complete metric spaces:Theorem 12 [9].
LetT be a mapping of a complete metric space X,d into itself satisfying
(53)dTx,Ty≤ϕmaxdx,Tx,dy,Ty,for all x,y∈X, where the function ϕ:0,∞⟶0,∞ is such that ϕt<t for each t>0. If ϕ is upper semicontinuous in the open interval 0,dTkX, then, T has a unique fixed point.Now, we will demonstrate an example to explain the above theorem:Example 6.
LetX=0,3 with usual metric dx,y=∣x−y∣ for all x,y∈X. Let a mapping T:X⟶X be defined by
(54)Tx=0,x∈0,11,x∈1,3.
Also, defineϕ:0,∞⟶0,∞ as
(55)ϕt=1+t2,t>1,t2,t≤1.
It is clear that the mappingT satisfies the criteria of Theorem 12 with a unique fixed point T=1 but it is discontinuous at this fixed point. Also, we observe that dTX=1 and ϕ is continuous on 0,1.Here, we present an extension of Theorem12 for a pair of maps to obtain a unique common fixed point.Theorem 13.
LetX be a nonempty set and let d be a metric on X. Let T and S be self mappings of a complete metric space X,d satisfying
(56)dTx,Sy≤ϕMx,y,for all x,y∈X, where the mapping ϕ:0,∞⟶0,∞ is such that ϕt<t for all t>0 and
(57)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2.
Ifϕ is upper semicontinuous on 0,dTkX and 0,dSkX for k=0,1,2,⋯, then, T and S have a unique common fixed point.Proof.
Letx0∈X. Define a sequence xninX as x2n+1=Tx2n and x2n+2=Sx2n+1, for all integers n≥0. If we assume that there exists a nonnegative integer n0 such that, x2n0=x2n0+1, then, x2n=x2n+1=Tx2n; this implies that x2n is a fixed point of T. Similarly, if there exists an integer N≥0 such that x2N+1=x2N+2, then, x2n+1 is a fixed point of S. This concludes the proof.☐Otherwise, we suppose thatx2n≠x2n+1, for all integers n≥0. Let μ2n=dx2n,x2n+1, obviously, μ2n+1=dx2n+1,x2n+2.From (84), we have
(58)dx2n+1,x2n+2=dTx2n,Sx2n+1≤ϕMx2n,x2n+1,where
(59)Mx2n,x2n+1=maxdx2n,x2n+1,dx2n,Tx2n,dx2n+1,Sx2n+1,dx2n,Sx2n+1+dx2n+1,Tx2n2,Using equations (8) and (12), we have the following:
(60)Mx2n,x2n+1=maxdx2n,x2n+1,dx2n+1,x2n+2.Thus,(61)dx2n+1,x2n+2=dTx2n,Sx2n+1≤ϕmaxdx2n,x2n+1,dx2n+1,x2n+2.If we takemaxdx2n,x2n+1,dx2n+1,x2n+2=dx2n+1,x2n+2, then,
(62)dx2n+1,x2n+2=dTx2n,Sx2n+1≤ϕdx2n+1,x2n+2<dx2n+1,x2n+2,which is a contradiction. Hence, max=dx2n,x2n+1,dx2n+1,x2n+2=dx2n,x2n+1.Therefore,
(63)dx2n+1,x2n+2=dTx2n,Sx2n+1≤ϕdx2n,x2n+1<dx2n,x2n+1,which implies that the sequence μ2n is decreasing to a nonnegative real number say δ, for all integers n≥0. We claim that δ=0. In contrary, suppose that δ>0. Taking the limit as n⟶∞ in (63), we obtain
(64)0<δ≤ϕδ<δ,which is a contradiction; hence, we conclude that δ=0 and
(65)limn⟶∞μ2n=limn⟶∞dx2n,x2n+1=0.Now, we need to show that a sequencexninX is a Cauchy sequence. We claim otherwise. Therefore, there exists ε>0 and a sequence of integers mr,nr such that
(66)dxnr,xmr≥ε,for all nr>mr≥r for some r≥0.Furthermore, suppose thatmr is the smallest integer which is chosen in such a way that (66) holds so that we have
(67)dxr,xmr−1<ε.Now, for allnr>mr, we have
(68)dxnr,xmr≤dxnr,xmr−1+dxmr−1,xmr≤dxnr,xmr−1+dxmr−1,xmr.Asr⟶∞ in (68) and considering (65) and (67), we see that
(69)dxnr,xmr⟶ε.By similar computations, we see that(70)dxnr−1,xmr−1⟶ε.Thus,(71)dxnr,xmr=dTxnr−1,Sxmr−1≤ϕMxnr−1,xmr−1,where
(72)Mxnr−1,xmr−1=maxdxnr−1,xmr−1,dxnr−1,Txnr−1,dxmr−1,Sxmr−1,dxnr−1,Sxmr−1+dxmr−1,Txnr−12=maxdxnr−1,xmr−1,dxnr−1,xnr−1dxmr−1,xmr−1,dxnr−1,xmr−1+dxmr−1,xnr−12,As r⟶∞ in (72) and considering (69) and (70), then, (71) becomes
(73)0<ε≤ϕε<ε,which is a contradiction. Hence, xninX is a Cauchy sequence and limn,m⟶∞dxn,xm=0.BecauseX is complete, we can pick a point z∈X such that limn⟶∞dx2n,z=0. Here, we will prove that z is a fixed point of S. In contrary, suppose that z≠Sz.Now,(74)dx2n+1,Sz=ϕdTx2n,Sz≤ϕMx2n,z.where
(75)Mx2n,z=maxdx2n,z,dx2n,Tx2n,dz,Sz,dx2n,Sz+dz,Tx2n2=maxdx2n,z,dx2n,Tx2n,dz,Sz,dx2n,Sz+dz,x2n+12.As n⟶∞ in (75), we see that
(76)Mx2n,z⟶dz,Sz.Applying the limit asn⟶∞ in (74), we have
(77)dz,Sz≤ϕdz,Sz<dz,Sz,which is a contradiction. Hence, Sz=z.Now, we will show that a pointz is a unique common fix of T and S. In contrary, suppose that z∈X and w∈X are two different common fixed points of T and S, respectively. Thus, dz,w>0.Now,(78)dz,z=dTz,Sw≤ϕMz,wwhere
(79)Mz,w=maxdz,w,dz,Tz,dw,Sw,dz,Sw+dw,Tz2=maxdz,w,dz,z,dw,w,dz,w+dw,z2=dz,w.Hence,(80)dz,w=dTz,Sw≤ϕdz,w<dz,w,which is a contradiction. Therefore, T and S have a unique common fixed point, that is z=w.To prove that any fixed point ofT is also a fixed point of S, conversely, we suppose to the contrary that z=Tz and z≠Sz.Now,(81)dTz,Sz=dz,Sz≤ϕMz,Swwhere
(82)Mz,Sw=maxdz,Sz,dz,Tz,dSz,S2z,dz,S2z+dSz,Tz2=maxdz,Sz,dz,z,dSz,S2z,dz,S2z+dSz,z2=dz,Sz.Thus,(83)dTz,Sz=dz,Sz≤ϕdz,Sz<dz,Sz,which is a contradiction. Therefore, z=Tz=Sz. In a similar way, it is easy to show that any fixed point of S is also a fixed point of T.On settingS=T, we get the following corollary:Corollary 14.
LetX be a nonempty set and let d be a metric on X. Let T be a self mapping of a complete metric space X,d satisfying
(84)dTx,Ty≤ϕMx,y,for all x,y∈X, where the mapping ϕ:0,∞⟶0,∞ is such that ϕt<t for all t>0 and
(85)Mx,y=maxdx,y,dx,Tx,dy,Ty,dx,Ty+dy,Tx2.
Ifϕ is upper semicontinuous on 0,d(TkX for k=0,1,2,⋯, then, T has a fixed point.Example 7.
LetX=0,3 be equipped with the usual metric. Let S and T be self mappings on X, i.e., T,S:X⟶X defined by
(86)Tx=1+x2,if0≤x≤1,0,ifx∈1,3,Sx=x2,if0≤x≤1,0,if1<x≤2,x2,if2<x≤3.
Defineϕ=1+t/2.Now, we have(87)dTx,Sy<ϕMx,y,where
(88)Mx,y=maxdx,y,dx,Tx,dy,Sy,dx,Sy+dy,Tx2,Forx,y≤1, using ϕ=1+t/2, we have
(89)dTx,Sy=x−2y2+12,Mx,y=x−12,dTx,Sy≤ϕMx,y,x−2y2+12≤ϕx−12,0≤1+x2,which is true.Forx,y>2, 2≤x≤3 and using ϕ=1+t/2, we have
(90)dTx,Sy=y2,Mx,y=x,dTx,Sy≤ϕMx,y,y2≤ϕx,y2≤1+x2.Forx≤1, y≤2 and using ϕ=1+t/2, we have
(91)dTx,Sy=x+12,Mx,y=y,dTx,Sy≤ϕMx,y,x+12≤ϕy,x+12≤1+y2,Hence, all conditions imposed in Theorem13 are satisfied. Thus, T and S satisfy all the conditions of the above theorem and has a unique fixed point x=1; S and T are discontinuous at x=1. The mapping T is 2-continuous and S is 3-continuous at x=1.


## 3. Fixed Points of Nonexpansive Mappings

Bisht and Pant [11] proved a fixed point theorem (see Theorem 15 [11]) for nonexpansive mapping. In this section, we are extending the result due to Bisht and Pant [11] for a pair of self mappings.In what follows, we shall denote(92)Px,y=maxdx,y,dx,Tx,dy,Sy,adx,Tx+dy,Sy2,bdx,Sy+dy,Tx2,0≤a,b<1.We will use this expression in the following theorem.Theorem 15.
LetX be a nonempty set and let d be a metric on X. Let T and S be self mappings of a complete metric space X,d such that for any x,y∈X(i)
For any givenε>0, there exists a δε>0 such that ε<Px,y<ε+δ implies dTx,Sy≤ε(ii)
dTx,Sy≤Px,y
Then,T and S have a unique common fixed point, say z and Tnx⟶z as well as Snx⟶z for each x∈X.Proof.
Letx0 be any point in X. Define a sequence xninX as x2n+1=Tnx0=Tx2n and x2n+2=Sx2n+1 for all integers n≥0. Then, on following the proof of Theorem 6, we can easily prove that xn is a Cauchy sequence in X. Since X is complete, there exists a point z∈X such that xn⟶z as n⟶∞. Also, Txn⟶z and Sxn⟶z as n⟶∞. We claim that Tz=z. For if Tz≠z, we get
(93)dTz,Sxn≤maxdz,xn,dz,Txn,dxn,Sxn,adz,Tz+dxn,Sxn2,bdz,Sxn+dxn,Tz2.
On lettingn⟶∞, this yields, dTz,Sz≤maxadTz,z+dz,Sz/2,bdSz,z+dz,Tz/2<maxdz,Tz,dz,Sz, which is a contradiction since 0≤a,b<1. Thus, z is a common fixed point of T and S.☐Remark 16.
By settingS=T, one can get Theorem 15 of Bist and Pant [11].Corollary 17.
LetX be a nonempty set and let d be a metric on X. Let T be a self mapping of a complete metric space X,d such that for any x,y∈X(i)
For any givenε>0, there exists a δε>0 such that ε<Px,y<ε+δ implies dTx,Ty≤ε(ii)
dTx,Ty≤Px,y
Then,T has a unique fixed point, say z and Tnx⟶z for each x∈X.Example 8.
LetX=−1,1 be equipped with usual metric and T,S:X⟶X be defined by
(94)Tx=0,ifx∈−1,0,−12,ifx∈0,1,Sx=x2,ifx∈−1,0,−14,ifx∈0,1.To verify our contraction condition, let us consider the following two cases:Case 1.
Forx,y∈−1,0, we have
(95)dx,y=x−y,dx,Tx=dx,0=x−0=x,dy,Sy=y,y2=y−y2=y2,dx,Sy=dx,y2=x−y2=2x−y2,dy,Tx=dy,0=y−0=y,dTx,Sy=d0,y2=y2.Thus, we have,(96)Px,y=maxdx,y,dx,Tx,dy,Sy,adx,Tx+dy,Sy2,bdx,Sy+dy,Tx2,Px,y=maxx−y,x,y2,ax+y/22,b2x−y/2+y2,Px,y=maxx−y,x,y2,a2x+y4,b2x+y4,=x−y,for0≤a,b≤1.Usingi and ii of Theorem 15, we have
(97)ε<Px,y<ε+δ⇒dTx,Sy≤ε,ε<x−y<ε+δ⇒y2≤ε,dTx,Sy≤Px,y⇒y2≤x−y.Case 2.
Forx,y∈0,1, we have
(98)dx,y=x−y=1−0=1,dx,Tx=d1,T0=1−−12=32,dy,Sy=0−S0=0−−14=14,dx,Sy=d1,S0=1−−14=54,dy,Tx=d0,T1=0−−12=12,dTx,Sy=dT1,S0=d−12,−14=14.Thus, we get(99)Px,y=dx,y,dx,Tx,dy,Sy,adx,Tx+dy,Sy2,bdx,Sy+dy,Tx2,P1,0=max1,32,14,a3/2+1/42,b5/4+1/22,Px,y=P1,0,=max1,32,14,78a,78b,=32,0≤a,b≤1.Finally, byi and ii of Theorem 15, we get
(100)ε<Px,y<ε+δ⇒dTx,Sy≤ε,ε<32<ε+δ⇒14≤ε,dTx,Sy≤Px,y⇒14≤32.Thus, we havedTx,Sy≤Px,y. Hence, the contraction condition of Theorem 15 is satisfied, and 0 is the common fixed point of T and S. Also, T is not continuous but is 2-continuous. Similarly, S is not continuous at 0 but is 2-continuous.


## 4. The Existence Solution of Nonlinear Volterra Integral Equation

The integral equation method is very useful for solving many problems in several applied fields like mathematical economics and optimal control theory because problems in these areas are often reduced to integral equations.Integral equations appear in several forms. However, in this section, we are interested in the integral equation, namely, the Volterra integral differential equation which is of the form(101)unt,x=ft,x+∫axKx,t,utdt,where un=dnu/dxn.Now, we present the application of Theorem6 to study the existence and uniqueness of the solution to nonlinear Volterra integral equations.The following integral equation is inspired by [14–17].
(102)ux,y=fx,y+∫0xgx,y,ε,uε,ydε+∫0x∫0yhx,y,σ,τ,uσ,τdεdσ,where f,g,h are given functions and u is the unknown function to be found.LetCT,S be the class of continuous functions from set T to set S. We denote E=ℝ+×ℝ+,E1=fx,y,s:0≤s≤x<∞,y∈ℝ+ and E2=fx,y,s,t:0≤s≤x<∞,0≤t≤y<∞. We denote that f∈CE,ℝ,g∈CE1×ℝ,ℝ, and h∈CE2×ℝ,ℝ.LetX be the space of functions z∈Cℝ+×ℝ+,ℝ satisfying
(103)zx,t=Oeλx+y,where λ is a positive constant, that is,
(104)zx,y≤M0eλx+y,for constant M0>0. Let X,. be a Banach space. Define a norm in the space X by
(105)zX=supx,y∈Xzx,ye−λx+y.Define the mappingT,S:X×X⟶0,∞ by
(106)Tkux,y=fx,y+∫0xgx,y,ε,uε,ydε+∫0x∫0yhx,y,σ,τ,uσ,τdεdσ,(107)Skux,y=fx,y+∫0xgx,y,ε,vε,ydε+∫0x∫0yhx,y,σ,τ,vσ,τdεdσ,for u,v∈X. We assume that Tk and Sk are k-continuous for some positive integer k. For sufficiently large values of k, the mappings Tk and Sk are contraction, where T and S are noncontraction if x−a>1, for x>a.Now, we prove our results by establishing the existence of a common fixed point for a pair of self mappings:Theorem 18.
Suppose that equation (102) satisfies the following conditions:
(i)
For the continuous functionsf,g∈X, we have(108)gx,y,ε,uε,y−gx,y,ε,vε,y≤L1x,y,εu−v,hx,y,σ,τ,uσ,τ−hx,y,σ,τ,vσ,τ≤L2x,y,σ,τu−v,where L1∈CE1,0,∞ and L2∈CE2,0,∞(ii)
There exist a nonnegative constantγ such that γ<1 and(109)∫0xL1x,y,εeλx+y+∫0x∫0yL2x,y,σ,τeλσ+τ≤δ,for all x,y,ε,σ,τ∈E1∪E2, and
(110)γ=λL1+L2eλx+y−L1λeλy−2L2eλx+L2λ2.
Then, the nonlinear Volterra integral equation (102) has a unique common solution in E1∪E2.Proof.
LetTk,Sk:X⟶X be two operators such that Tk∈X and Sk∈X. Now, we verify that Tk and Sk are contractive maps in X. Let u,v∈X. On the contrary, we claim that neither Tk nor Sk are contractive maps in X. From (106) and (107), using condition i and ii of Theorem 18, we have
(111)Tku−Skv=fx,y+∫0xgx,y,ε,uε,ydε+∫0x∫0yhx,y,σ,τ,uσ,τdεdσ−fx,y−∫0xgx,y,ε,vε,ydε−∫0x∫0yhx,y,σ,τ,vσ,τdεdσ,≤∫0tgx,y,ε,uε,y−gx,y,ε,vε,ydε+∫0x∫0yhx,y,σ,τ,uσ,τ−hx,y,σ,τ,vσ,τ,≤∫0xL1x,y,εeλx+y+∫0x∫0yL2x,y,σ,τeλσ+τ∥u−v∥X,≤1λL1eλx+y−eλy+1λ2L2eλx+y−2eλx+1∥u−v∥X,≤λL1+L2eλx+y−L1λeλy−2L2eλx+L2λ2∥u−v∥X,⇒Tku−Skv≤γ∥u−v∥X⇒dTu,Sv≤γMu,v,which is a contradiction. Hence, u is a common fix of T and S and also a solution to integral equation (102).
From (111), let γ=1 and using i of Theorem 6, where
(112)Mu,v=maxdu,v,du,Tv,dv,Sv,du,Sv+dv,Tu2,we have
(113)dTu,Sv<Mu,v.
Thus, Theorem6 is satisfied.☐



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*Source: 9982217-2021-08-16.xml*