# Fixed Point Theorems in Complete Cone Metric Spaces over Banach Algebras

Authors: Seong-Hoon Cho
Journal: Journal of Function Spaces
 (2018)

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## Abstract

The notion of C-class functions in Banach algebras is introduced. By using such concept, a new fixed point theorem is established. An example to illustrate main theorem is given. Finally, applications of our main result to cyclic mappings and weak contraction type mappings are given.


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## Body

## 1. Introduction and Preliminaries

Huang and Zhang [1] introduced cone metric spaces which are generalizations of metric spaces, and they extended Banach’s contraction principle to such spaces, whereafter many authors (for example, [2–6] and references therein) studied fixed point theorems in cone metric spaces.Recently, Liu and Xu [7] introduced the notion of cone metric spaces over Banach algebras, which is a modification of the concept of cone metric spaces over real Banach spaces [1], and proved the existence of fixed points for mappings defined on such spaces, and they gave an example that fixed point results in metric spaces and in cone metric spaces are not equivalent.Very recently, Chandok et al. [8] introduced the concept of TAC-contractive mappings by using the notions of C-class functions and cyclic (α,β)-admissible mappings and established corresponding fixed point theorems in metric spaces.In the paper, we introduce the notions of C-class functions and cyclic (α,β)-admissible mappings in Banach algebras. By using such concepts, we introduce a new contraction. We obtain a new fixed point theorem and give an example to illustrate main result. Finally, we give applications of our main result to cyclic mappings and weak contraction type mappings in cone metric spaces over Banach algebras.A Banach spaceA is called a (real) Banach algebra (with unit) if there exists a multiplication A×A→A that has the followings properties:for allx,y,z∈A,α∈R,(1)
(xy)z=x(yz);(2)
x(y+z)=xy+xz and (x+y)z=xz+yz;(3)
α(xy)=(αx)y=x(αy);(4)
there existse∈A such that xe=ex=x;(5)
e=1;(6)
xy≤xy.An elementx∈A is calledinvertible if there exists x-1∈A such that xx-1=x-1x=e.Proposition 1 (see [9]).
LetA be a Banach algebra, and let x∈A. If the spectral radius ρ(x) of x is less than 1, i.e., (1)ρx=limn→∞xn1/n=infn≥1xn1/n<1,then e-x∈G(A), where G(A) is the set of all invertible elements of A and (2)e-x-1=∑n=0∞xn.Remark 2.
LetA be a Banach algebra.Then the following are satisfied:(1)
for allx∈A, ρ(x)≤x;(2)
if the conditionρ(x)<1 is replaced by x<1 in Proposition 1, then the conclusion holds.Consistent with Liu and Xu [7], the following definitions will be needed in the sequel.LetA be a Banach algebra. A subset P of A is calledcone if the following conditions are satisfied:(1)
P is a nonempty and closed subset of A and {0,e}⊂P;(2)
ax+by∈P, whenever x,y∈P and a,b∈R(a,b≥0);(3)
P2=PP⊂P;(4)
P∩(-P)={0}.Remark 3.
LetA be a Banach algebra, and P⊂A be a cone.
Ifx∈P and ρ(x)<1, then e-x∈G(P).Given a coneP⊂A, we define a partial ordering ≤ with respect to P by x≤y if and only if y-x∈P. We write x<y to indicate that x≤y but x≠y.Forx,y∈P, x≪y stand for y-x∈int(P), where int(P) is the interior of P. A cone P is callednormal if there exists a number K≥1 such that, for all x,y∈A, x≤Ky whenever 0≤x≤y.A coneP is calledregular if every increasing sequence which is bounded from above is convergent. That is, if {un} is a sequence such that for some z∈A(3)u1≤u2≤⋯≤z,then there exists u∈A such that (4)limn→∞un-u=0.Equivalently, a coneP is regular if and only if every decreasing sequence which is bounded from below is convergent.It is well known that every regular cone is normal.From now on, we assume thatA is a Banach algebra, P⊂A is a cone with int(P)≠∅, and ≤ is partial ordering with respect to P.LetX be a nonempty set, and let d:X×X→A be a map such that(1)
0≤d(x,y) for all x,y∈X and d(x,y)=0 if and only if x=y;(2)
d(x,y)=d(y,x) for all x,y∈X;(3)
d(x,y)≤d(x,z)+d(z,y) for all x,y,z∈X.Thend is called a cone metric on X (with a Banach algebra A) and (X,d) is called cone metric space (with a Banach algebra A).Remark 4.
IfA=C and P={z∈C:Re(z)≥0andIm(z)≥0}, where C is the set of all complex numbers, then we have that (1)
z1≤z2⇔[Re(z1)≤Re(z2)andIm(z1)≤Im(z2)]⇔z2-z1∈P;(2)
the complex valued metric [10] is a cone metric.A sequence{xn} of points in a cone metric space (X,d)converges to a point x∈X (denoted by limn→∞xn=x or xn→x) if for any c∈A with 0≪c, there exists N such that for all n>N, d(xn,x)≪c. A sequence {xn} of points in a cone metric space (X,d) isCauchy if for any c∈A with 0≪c, there exists N such that for all n,m>N, d(xn,xm)≪c. A cone metric space (X,d) is calledcomplete if every Cauchy sequence is convergent.Remark 5.
Let(X,d) be a cone metric space and {xn}⊂X be a sequence and x∈X. Then the following are satisfied: (1)
limn→∞xn=x if and only if for any c∈int(P), there exists N such that for all n>N, c-d(xn,x)∈int(P);(2)
{xn} is a Cauchy sequence if and only if for any c∈int(P), there exists N such that for all n,m>N, c-d(xn,xm)∈int(P).Note that iflimn→∞d(xn,x)=0, then limn→∞xn=x. The converse is true if P is a normal cone. Also, if P is a normal cone, then {xn} is a Cauchy sequence in X if and only if limn,m→∞d(xn,xm)=0.Let(X,d) be a cone metric space.LetΦ be the class of all continuous function ϕ:int(P)∪{0}→int(P)∪{0} satisfying the following:(ϕ1) ϕ-1({0})=0;(ϕ2) ϕ(t)≪t for all t∈int(P);(ϕ3) either ϕ(t)≤d(x,y) or d(x,y)≪ϕ(t) for all t∈int(P) and x,y∈X.Lemma 6 (see [11]).
Let(X,d) be a cone metric space with regular cone P such that d(x,y)∈int(P) for x≠y. Let {xn}⊂X be a sequence, and let ϕ∈Φ.
If the sequence{xn} is not a Cauchy sequence, then there exist c∈int(P) and two subsequences {xm(k)} and {xn(k)} of {xn} such that m(k) is the smallest index for which m(k)>n(k)>k, (5)ϕc-dxmk,xnk∉intPand (6)ϕc-dxmk-1,xnk∈intP
Moreover iflimn→∞d(xn,xn+1)=0, then we have (1)
limn→∞d(xm(k),xn(k))=ϕ(c);(2)
limn→∞d(xm(k)+1,xn(k)+1)=ϕ(c).Let(X,d) be a cone metric space.A mapT:X→X is calledcontinuous at x∈X if for any V∈τ containing Tx, there exists U∈τ containing x such that TU⊂V, where τ is the topology on X induced by the cone metric d. That is,τ = { U ⊂ X : ∀ x ∈ U , ∃ B ∈ β , x ∈ B ⊂ U },β = { B ( x , c ) : x ∈ X , ∀ c ∈ i n t ( P ) },B ( x , c ) = { y ∈ X : d ( x , y ) ≪ c }.IfT is continuous at each point x∈X, then it is calledcontinuous.Note thatT is continuous if and only if it is sequentially continuous, i.e., limn→∞Txn=Tx for any sequence {xn}⊂X with limn→∞xn=x (see [12]).A pointx∈X is calledcluster point of A⊂X if for every c∈A with 0≪c, (7)Bx,c-x∩A≠∅.LetΨ be the family of all continuous function ψ:P→P such that(ψ1) ψ is strictly increasing, i.e., x<y⇔ψ(x)<ψ(y);(ψ2) ψ-1({0})=0.Note that ifψ(x)≤ψ(y), then x≤y.Alizadeh et al. [13] introduced the concept of cyclic (α,β)-admissible mappings in R+.We extend this concept to cones as follows.Letα,β:X→P be functions, where X is a set. We say that T:X→X is acyclic (α,β)-admissible mapping if(1)
αx-e∈P,wherex∈X⇒β(Tx)-e∈P;(2)
β(x)-e∈P,wherex∈X⇒α(Tx)-e∈P.Example 7.
LetK be a compact Hausdorff topological space, and let A be the family of all continuous function from K into R.
Define the multiplicationxy of x and y as follows: (8)xy=xtyt∀t∈K.
ThenA is a Banach algebra with unit e(t)=1∀t∈K.
LetP={x∈A:x(t)≥0∀t∈K}.
ThenP is a normal cone with normal constant M=1.
LetX=A, and let α,β:X→P be functions defined by (9)αx=βx=x∨e.
DefineT:X→X as follows: (10)Tx=x∨2.
Then we have(11)αx-e∈P,x∈P⇒βTx-e=x∨2-e∈Pand (12)βx-e∈P,x∈P⇒αTx-e=x∨2-e∈P.
HenceT is cyclic (α,β)-admissible.


## 2. Fixed Points

LetC be a family of all continuous functions f:P×P→P such that(1)
f(s,t)≤s∀s,t∈P;(2)
f(s,t)=s⇒ either s=0 or t=0∀s,t∈P.Then we say thatf∈C is C-class function.Note thatf(0,0)=0.Example 8.
LetA be a Banach algebra, and let P⊂A be a cone.(1)
Letfb(s,t)=ks, where k∈P-{0} with ρ(k)<1.Sincek∈P-{0} and ρ(k)<1, from Remark 4e-k∈G(P). Hence s-ks=s(e-k)∈P∀s∈P, and hence 0≤s-ks. So ks≤s∀s∈P. Thus fb(s,t)=ks≤s∀s,t∈P.Iffb(s,t)=s, then ks=s, and so s(e-k)=0∈P. Since e-k∈G(A), e-k≠0. Hence s=0. Thus fb∈C.(2)
Letfbw(s,t)=φ(s), where φ:P→P is a continuous function such that φ-1({0})=0 and φ(s)<s∀t∈P-{0}.Thenfbw(s,t)=φ(s)≤s∀s,t∈P.Letfbw(s,t)=s.Thens=φ(s). If s≠0, then s=φ(s)<s, which is a contradiction.Hences=0. Thus fbw∈C.(3)
Letfw(s,t)=s-ϕ(s), where ϕ∈Φ.Thenfw(s,t)≤s∀s,t∈P.Iffw(s,t)=s, then ϕ(s)=0, and so s=0. Hence fw∈C.(4)
Letfr(s,t)=sβ(s)∀s,t∈P,whereβ:P×P→B={x∈P-{0}:ρ(x)<1} is continuous.Sinceβ(s)∈P-{0} and ρ(β(s))<1, e-β(s)∈G(P). Hence s-β(s)s=s(e-β(s))∈P∀s∈P, and hence β(s)s≤s∀s∈P. Thus fr(s,t)=sβ(s)≤s∀s,t∈P.Iffr(s,t)=s, then s(e-β(s))=s-sβ(s)=0∈P. Since e-β(s)∈G(P), e-β(s)≠0. Hence s=0.(5)
Letfh(s,t)=sh(s,t)∀s,t∈P,whereh:P×P→B={x∈P-{0}:ρ(x)<1} is continuous.Sinceh(s,t)∈P-{0} and ρ(h(s,t))<1, e-h(s,t)∈G(P). Hence s-h(s,t)s=s(e-h(s,t))∈P, and hence sh(s,t)≤s. Thus fh(s,t)=sh(s,t)≤s∀s,t∈P.Iffh(s,t)=s, then s(e-h(s,t))=s-sh(s,t)=0∈P.We havee-h(s,t)≠0, because e-h(s,t)∈G(P).Hences=0, Thus fh∈C.From now on, letX be a cone metric space with cone metric d and regular cone P such that for x,y∈X with x≠y(13)dx,y∈intP.Theorem 9.
LetX be complete, and let T:X→X be a mapping such that (14)ψdTx,Ty≤fψdx,y,ϕdx,yfor all x,y∈X with α(x)β(y)-e∈P, where f∈C, ψ∈Ψ, and ϕ∈Φ.
Also, suppose that the following are satisfied:(1)
T is cyclic (α,β)-admissible;(2)
There existsx0∈X such that α(x0)-e∈P and β(x0)-e∈P;(3)
EitherT is continuous or if {xn}⊂X is a sequence with β(xn)-e∈P∀n=1,2,3,⋯ and x is a cluster point of {xn}, then (15)βx-e∈P.
ThenT has a fixed point.
Moreover, if(16)αx-e∈Pandβy-e∈Pfor all fixed points x,y of T, then T has a unique fixed point.Proof.
Define a sequence{xn} in X by xn=Txn-1=Tnx0 for all n=1,2,3,⋯.
If there exists an integerN such that xN=xN+1, then xN=TxN; i.e., xN is a fixed point of T.
Hence we assume thatxn≠xn+1∀n=1,2,3,⋯.
Sinceα(x0)-e∈P, from (1) we have β(x1)-e=β(Tx0)-e∈P. Again from (1) we obtain α(x2)-e=α(Tx1)-e∈P.
Inductively, we have(17)αx2n-e∈Pandβx2n+1-e∈P∀n=0,1,2,⋯.
SinceT is cyclic (α,β)-admissible and β(x0)-e∈P, α(x1)-e=α(Tx0)-e∈P. Similarly, we obtain (18)βx2n-e∈Pandαx2n+1-e∈P∀n=0,1,2,⋯.
Hence(19)αxn-e∈Pandβxn-e∈P∀n=0,1,2,⋯.
Sinceα(xn+1)-e∈P and β(xn)-e∈P, (20)αxn+1βxn-αxn+1-βxn+e=αxn+1-eβxn-e∈P.
Also,(21)αxn+1+βxn-2e=αxn+1-e+βxn-e∈P.
Hence,(22)αxn+1βxn-e∈P∀n=1,2,3,⋯And so it follows from (14) that(23)ψdxn,xn+1=ψdTxn-1,Txn≤fψdxn-1,xn,ϕdxn-1,xn≤ψdxn-1,xn∀n=1,2,3,⋯.
Sinceψ is strictly increasing, (24)dxn,xn+1≤dxn-1,xn∀n=1,2,3,⋯.
Hence{d(xn-1,xn)} is a decreasing sequence.
SinceP is regular, there exists r∈P such that (25)limn→∞dxn-1,xn=r.By Lemma 3.1 of [14], we have (26)r∈intP∪0.
Assume thatr∈int(P).
By taking limitn→∞ in (23) and using continuity of f,ψ, and ϕ, we obtain (27)ψr≤fψr,ϕr≤ψrwhich implies (28)ψr-fψr,ϕr∈P∩-P.So (29)fψr,ϕr=ψr,and so either ψ(r)=0 or ϕ(r)=0. Hence r=0, which is a contradiction.
Hence(30)r∉intP.
Thus we have(31)limn→∞dxn,xn-1=0.
We now show that{xn} is a Cauchy sequence.
On the contrary, assume that{xn} is not a Cauchy sequence.
By Lemma6, there exists c∈int(P) and two subsequences {xm(k)} and {xn(k)} of {xn} such that m(k) is the smallest index for which, for all k∈N, m(k)>n(k)>k, (5) and (6) hold.
It follows from (19) that α(xm(k))β(xn(k))-e∈P, and so from (14) with x=xm(k) and y=xn(k) we have(32)ψdxmk+1,xnk+1=ψdTxmk,Txnk≤fψdxmk,xnk,ϕdxmk,xnk.
By Lettingk→∞ and by using Lemma 6 (1) and (2), and using continuity of ψ,ϕ, and f, we have (33)ψϕc≤fψϕc,ϕϕc≤ψϕcwhich implies (34)ψϕc-fψϕc,ϕϕc∈P∩-P.
Hence(35)fψϕc,ϕϕc=ψϕcwhich implies either ψ(ϕ(c))=0 or ϕ(ϕ(c))=0. Thus ϕ(c)=0, and so c=0, which is a contradiction.
Therefore,{xn} is a Cauchy sequence.
It follows from the completeness ofX that there exists (36)x∗=limn→∞xn∈X.
IfT is continuous, then limn→∞xn=Tx∗, and so x∗=Tx∗.
Assume that (15) holds.
Then we have(37)βx∗-e∈P.
SinceT is (α,β)-admissible, α(xn)-e∈P∀n=1,2,3,⋯.
Using (19) we have (38)αxnβx∗-e∈P∀n=1,2,3,⋯.
From (14) we have(39)ψdxn+1,Tx∗=ψdTxn,Tx∗≤fψdxn,x∗,ϕdxn,x∗.
Lettingn→∞ in the inequality (39) and using continuity of f,ψ, and ϕ, we have (40)ψdx∗,Tx∗≤fψ0,ϕ0=0,andsoψdx∗,Tx∗∈-Pwhich implies (41)ψdx∗,Tx∗∈P∩-P.
Thus(42)ψdx∗,Tx∗=0.Hence (43)dx∗,Tx∗=0,andhencex∗=Tx∗.
For the uniqueness of fixed point, assume that (16) holds and that y∗ is another fixed point of T.
Then from (16) we have α(x∗)-e∈P and β(y∗)-e∈P.
Henceα(x∗)β(y∗)-e∈P.
It follows from (14) that (44)ψdx∗,y∗=ψdTx∗,Ty∗≤fψdx∗,y∗,ϕdx∗,y∗≤ψdx∗,y∗which implies (45)ψdx∗,y∗-fψdx∗,y∗,ϕdx∗,y∗∈P∩-P.
Thus(46)fψdx∗,y∗,ϕdx∗,y∗=ψdx∗,y∗which implies (47)eitherψdx∗,y∗=0orϕdx∗,y∗=0.So d(x∗,y∗)=0, and hence x∗=y∗.Example 10.
LetA=R2,P={(x1,x2)∈A:x1,x2≥0}, and let x=x1+x2∀x=(x1,x2)∈A.
Define the multiplicationxy of x=(x1,x2) and y=(y1,y2) as follows: (48)xy=x1,x2y1,y2=x1y1,x1y2+x2y1.
ThenA is a Banach algebra with unit e=(1,0), P is regular cone, and int(P)={(x1,x2)∈A:x1,x2>0}.
LetX=R2, and let d:X×X→A be a mapping defined by (49)dx,y=max⁡x1-y1,x2-y2.
Then(X,d) is a complete cone metric space, and d(x,y)∈int(P)∀x,y∈Xwithx≠y.
Define a mappingT:X→X by (50)Tx=14x1,x2,x=x1,x2∈Pandx<1;logx1x2,ex1x2,otherwise.
Letφ(t)=(1/2)t, ψ(t)=t, and ϕ(t)=(1/3)t∀t=(t1,t2)∈P.We define two functionsα,β:X→P by (51)αx=βx=ex=x1,x2∈Pandx<1;0otherwise.Obviously,T is (α,β)-admissible.Letf(s,t)=fbw(s,t)∀s,t∈P.Then we have(52)fbwψdx,y,ϕdx,y-ψdTx,Ty=fbwdx,y,ϕdx,y-dTx,Ty=φdx,y-dTx,Ty=12maxx1-y1,x2-y2-d14x1,x2,14y1,y2=12maxx1-y1,x2-y2-14maxx1-y1,x2-y2=14maxx1-y1,x2-y2∈P∀x,y∈Xwithαxβy-e∈P.Henceψ(d(Tx,Ty))≤fbw(ψ(d(x,y)),ϕ(d(x,y)))∀x,y∈X with α(x)β(y)-e∈P.Condition (2) of Theorem9 is satisfied with x0=(0,0).We now show condition (15) of Theorem 9.Letxn=(1/(n+4),2/(n+4))∀n=1,2,3,⋯.Thenxn∈P and xn<1∀n=1,2,3,⋯. Hence (53)βxn-e=0,0∈P∀n=1,2,3,⋯.Then limn→∞xn=(0,0), and so (0,0) is a cluster point of {xn}.Then we obtain(54)β0,0-e=0,0∈P.Hence condition (15) of Theorem 9 is satisfied.Thus, all hypotheses of Theorem9 are satisfied and T has a fixed point x∗=(0,0).Corollary 11.
LetX be complete, and let T:X→X be a mapping such that (55)ψαxβydTx,Ty≤fψdx,y,ϕdx,yfor all x,y∈X, where f∈C, ψ∈Ψ, and ϕ∈Φ.
Also, suppose that conditions (1), (2), and (3) of Theorem9 hold.
ThenT has a fixed point.
Moreover, if (16) holds, then T has a unique fixed point.


## 3. Discussion

We give an application of Theorem9 to prove existence of fixed points for cyclic mappings.Theorem 12.
LetX be complete, and let A and B be two closed subsets of X such that A∩B≠∅, and let T:A∪B→A∪B be a mapping such that (56)TA⊂BandTB⊂A.
Assume that(57)ψdTx,Ty≤fψdx,y,ϕdx,yfor all x∈A and y∈B, where f∈C, ψ∈Ψ, and ϕ∈Φ.
ThenT has a unique fixed point in A∩B.Proof.
Defineα,β:X→P by (58)αx=e,x∈A0,otherwise,(59)βx=e,x∈B0,otherwise.
Then we have that∀x,y∈A∪B(60)αxβy-e∈P⇔e≤αxβy⇔x∈Aandy∈B.
Hence (57) implies (13).
It follows from (56), (58), and (59) that (61)αx-e∈P⇒e≤αx⇒x∈A⇒Tx∈B⇒e≤βTx⇒βTx-e∈Pand (62)βx-e∈P⇒e≤βx⇒x∈B⇒Tx∈A⇒e≤αTx⇒αTx-e∈P.
ThusT is cyclic (α,β)-admissible.
BecauseA∩B≠∅, there exists an x0 in A∩B. By definition of functions α and β, α(x0)-e∈P and β(x0)-e∈P. Hence condition (2) of Theorem 9 is satisfied.Let{xn}⊂X be a sequence such that (63)limn→∞dxn,x=0andβxn-e∈P∀n=1,2,3,⋯.Thenxn∈B,∀n=1,2,3,⋯. Since B is closed, x∈B and so β(x)-e∈P. Hence condition (3) of Theorem 9 holds.Thus all conditions of Theorem9 are satisfied. It follows from Theorem 9 that T has a fixed point in A∪B; say x∗∈A∪B.Ifx∗∈A, then from (56) x∗=Tx∗∈B. Thus x∗∈A∩B. Similarly, we have x∗∈A∩B, when x∗∈B.Letx,y be any two fixed point of T.Thenx,y∈A∩B, and so α(x)β(y)-e∈P. It follows from Theorem 9 that T has a unique fixed point in A∩B.Corollary 13.
LetX be complete, and let A and B be two closed subsets of X such that A∩B≠∅, and let T:A∪B→A∪B be a mapping such that (64)TA⊂BandTB⊂A.
Assume that(65)dTx,Ty≤fdx,y,ϕdx,yfor all x∈A and y∈B, where f∈C and ϕ∈Φ.
ThenT has a unique fixed point in A∩B.Corollary 14.
LetX be complete, and let A and B be two closed subsets of X such that A∩B≠∅, and let T:A∪B→A∪B be a mapping such that (66)TA⊂BandTB⊂A.
Assume that(67)dTx,Ty≤kdx,yfor all x∈A and y∈B, where k∈P-{0} and ρ(k)<1.
ThenT has a unique fixed point in A∩B.Proof.
From (67) we have (68)dTx,Ty≤kdx,y⇔dTx,Ty≤fb(dx,y,ϕdx,yfor all x∈A and y∈B, where k∈P-{0}, ρ(k)<1, and ϕ∈Φ.
By Corollary13 with f(s,t)=fb(s,t)∀s,t∈P, T has a unique fixed point in A∩B.Remark 15.
IfA=R and P=[0,∞), then Corollary 14 reduces to Theorem 1.1 of [15].Remark 16.
Corollary14 is a generalization of Theorem 2.1 of [7] (resp., Theorem 3.1 of [16]) to cyclic mappings under assumption of regularity.


## 4. Applications

By applyingC-class functions of Example 8 to Theorem 9, we derive some existing fixed point results in the literature and have generalizations of well-known fixed point theorems in metric spaces to cone metric spaces.In particular, by takingf(s,t)=fw(s,t), we have the following result.Theorem 17.
LetX be complete, and let T:X→X be a cyclic (α,β)-admissible mapping such that (69)ψdTx,Ty≤ψdx,y-ϕdx,yfor all x,y∈X with α(x)β(y)-e∈P, where ψ∈Ψ and ϕ∈Φ such that (70)ϕψt≤ϕt∀t>0.
Suppose thatα(x0)-e∈P and β(x0)-e∈P, where x0∈X.
Assume that either(1)
T is continuous or(2)
if{xn}⊂X is a sequence such that limn→∞d(xn,x)=0 and β(xn)-e∈P∀n=1,2,3,⋯, then (71)βx-e∈P.
ThenT has a fixed point in X. Further if α(x)β(y)-e∈P for all fixed points x,y of T, then T has a unique fixed point.Proof.
Letfw(s,t)=s-ϕ(s)∀t,s∈P, where ϕ∈Φ.
Thenfw is a C-class function.
From (69) and (70) we have that for all x,y∈X with α(x)β(y)-e∈P(72)0≤ψdx,y-ϕdx,y-ψdTx,Ty≤ψdx,y-ϕψdx,y-ψdTx,Ty=fwψdx,y,ϕdx,y-ψdTx,Ty.
Hence(73)ψdTx,Ty≤fwψdx,y,ϕdx,yfor all x,y∈X with α(x)β(y)-e∈P.
It follows that (13) holds with f=fw. All conditions of Theorem 9 hold, and so we have a desired result.Corollary 18.
LetX be complete, and let T:X→X be a cyclic (α,β)-admissible mapping such that (74)dTx,Ty≤dx,y-ϕdx,yfor all x,y∈X with α(x)β(y)-e∈P, where ϕ∈Φ.
Suppose thatα(x0)-e∈P and β(x0)-e∈P, where x0∈X.
Assume that either(1)
T is continuous or(2)
if{xn}⊂X is a sequence such that limn→∞d(xn,x)=0 and β(xn)-e∈P∀n=1,2,3,⋯, then (75)βx-e∈P.
ThenT has a fixed point in X. Further if α(x)β(y)-e∈P for all fixed points x,y of T, then T has a unique fixed point.
By takingα(x)=e and β(x)=e∀x∈X in Theorem 17, we have the following result.Corollary 19.
LetX be complete, and let T:X→X be a mapping such that (76)ψdTx,Ty≤ψdx,y-ϕdx,yfor all x,y∈X, where ψ∈Ψ and ϕ∈Φ with ϕ(ψ(t))≤ϕ(t)∀t>0.
ThenT has a unique fixed point in X.Corollary 20.
LetX be complete, and let T:X→X be a mapping such that (77)dTx,Ty≤dx,y-ϕdx,yfor all x,y∈X, where ϕ∈Φ.
ThenT has a unique fixed point in X.Remark 21.
By takingA a real Banach space in Corollary 19 (resp., Corollary 20), we have Theorem 2.1 of [10] (resp., Theorem 2.1 of [4]).



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*Source: 9395057-2018-08-05.xml*