Common fixed points of two pairs of selfmaps satisfying a contractive condition with rational expression

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Available online at http://scik.org

Adv. Fixed Point Theory, 7 (2017), No. 2, 243-265 ISSN: 1927-6303

COMMON FIXED POINTS OF TWO PAIRS OF SELFMAPS SATISFYING A CONTRACTIVE CONDITION WITH RATIONAL EXPRESSION

G. V. R. BABU, K. K. M. SARMA AND D. RATNA BABU∗

Department of Mathematics, Andhra University, Visakhapatnam-530 003, India

Copyright c2017 G. V. R. Babu, K. K. M. Sarma and D. Ratna Babu. 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. Abstract. In this paper, we prove the existence of common fixed points of two pairs of selfmaps under the assumptions that these two pairs of maps are weakly compatible and satisfying a contractive condition involving rational expression in a complete metric space. The same is extended to a sequence of selfmaps. Also, we prove the same with different hypotheses on two pairs of selfmaps in which one pair is compatible, reciprocally continuous and the other one is weakly compatible. We also discuss the importance of rational expression in our contractive condition. Our theorems extend the results of Chandok [1] to two pairs of selfmaps.

Keywords: common fixed points; complete metric space; compatible maps; weakly compatible maps and recip-rocally continuous maps.

2010 AMS Subject Classification:47H10, 54H25.

1. Introduction

The development of fixed point theory is based on the generalization of contraction conditions in one direction or/and generalization of ambiant spaces of the operator under consideration on

∗_{Corresponding author}

E-mail address: ratnababud@gmail.com Received December 28, 2016

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the other. Banach contraction principle plays an important role in solving nonlinear equations, and it is one of the most useful results in fixed point theory. Banach contraction principle has been generalized in various ways either by using contractive conditions or by imposing some additional conditions on the ambiant spaces. In the direction of generalization of contraction conditions, in 1975, Dass and Gupta [2] established fixed point results using contraction conditions involving rational expressions.

The following theorem is due to Dass and Gupta [2].

Theorem 1.1. [2] Let(X,d)be a complete metric space and T :X →X a mapping such that there existα,β >0 withα+β <1 satisfying

d(T x,Ty)≤αd(x,y) +βd(y,Ty)[1+d(x,T x)]

1+d(x,y)

for allx,y∈X.ThenT has a unique fixed point.

Definition 1.1. [3] Let A and Bbe selfmaps of a metric space (X,d). The pair(A,B) is said to be a compatible pair onX, if lim

n→∞

d(ABxn,BAxn) =0 whenever{xn}is a sequence inX such

that lim

n→∞

Ax_n= lim

n→∞

Bx_n=t, for somet ∈X.

Definition 1.2. [4] LetAandBbe selfmaps of a metric space(X,d). The pair(A,B)is said to be weakly compatible, if they commute at their coincidence points. i.e.,ABx=BAxwhenever

Ax=Bx,x∈X.

Every compatible pair of maps is weakly compatible, but its converse need not true [4]. Definition 1.3. [5] LetAandBbe selfmaps of a metric space(X,d). ThenAandBare said to be reciprocally continuous, if lim

n→∞

ABx_n=At and lim

n→∞

BAx_n=Bt, whenever{x_n}is a sequence inX such that lim

n→∞

Axn= lim n→∞

Bxn=t, for somet ∈X.

Clearly, ifAandBare continuous then they are reciprocally continuous but its converse need not be true [5].

Recently, Chandok [1] established the following common fixed point result of selfmaps satisfying certain contraction condition involving rational expression.

Theorem 1.2. (Theorem 2.1, [1]) Let M be a subset of a metric space(X,d). Suppose that

T,f,g:M→M satisfy

d(T x,f y)≤α( d(gx,T x)d(gy,f y)

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for allx,y∈Mand for someα,β ∈[0,1)withα+β <1.

Suppose also thatT(M)∪f(M)⊆g(M)and(g(M),d)is complete.

ThenT,f andghave a coincidence point inM. Also, if the pairs(g,T)and(g,f)are weakly compatible, thenT,f andghave a unique common fixed point inX.

Throughout this paper, we denoteR+= [0,∞)and

Ψ={ψ/ψ:R+ →R+ is continuous,ψ is nondecreasing,ψ(t)<t fort>0 and ψ(t) =0⇔t=0}.

In Section 2, we prove the existence of common fixed points for two pairs of selfmaps under the assumptions that these two pairs of maps are weakly compatible and satisfying a contractive condition involving rational expression in a complete metric space. Also, we prove the same with different hypotheses on two pairs of selfmaps in which one pair is compatible, reciprocally continuous and the other one is weakly compatible. In section 3, we draw some corollaries from our main results and provide examples in support of our results and discuss the importance of rational expression in our contractive condition (Example 2).

2. Main results

LetA,B,SandT be mappings from a metric space(X,d)into itself and satisfying

(1) A(X)⊆T(X)andB(X)⊆S(X).

Now by (A), for anyx0∈X, there existsx1∈X such thaty0=Ax0=T x1. In the same way for thisx₁, we can choose a pointx₂∈X such thaty₁=Bx₁=Sx₂ and so on. In general, we can define a sequence{y_n}inX such that

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Lemma 2.1. Let(X,d)be a metric space. Assume thatA,B,SandT are selfmaps ofX which satisfy the following condition: there existsψ ∈Ψsuch that

(3) d(Ax,By)≤

      

     

ψ(max{_d₍_Sx_,_Tyd(₎₊Sx_d,Ax₍_Sx)d_,_By(Ty₎₊,By_d₍)_Ty_,_Ax₎,d(Sx,Ty)}),

if d(Sx,Ty) +d(Sx,By) +d(Ty,Ax)6=0 0, if d(Sx,Ty) +d(Sx,By) +d(Ty,Ax) =0 for allx,y∈X. Then we have the following:

(i) IfA(X)⊆T(X)and the pair(B,T)is weakly compatible, and ifzis a common fixed point ofAandSthenzis a common fixed point ofA,B,SandT and it is unique.

(ii) IfB(X)⊆S(X)and the pair(A,S)is weakly compatible, and ifzis a common fixed point ofBandT thenzis a common fixed point ofA,B,SandT and it is unique.

Proof.First, we assume that(i)holds. Letzbe a common fixed point ofAandS. ThenAz=Sz=z.SinceA(X)⊆T(X), there existsu∈X such thatTu=z. ThereforeAz=Sz=Tu=z.We now prove thatAz=Bu.Suppose thatAz6=Bu. We consider,

d(Az,Bu)≤ψ(max{_d₍_Sz_,_Tud(₎₊Sz,_dAz₍_Sz)d_,_Bu(Tu₎₊,Bu_d₍)_Tu_,_Az₎,d(Sz,Tu)}) =ψ(max{0,0}) =ψ(0) =0.

Therefore,Az=Bu.

HenceAz=Bu=Sz=Tu=z.

Since the pair(B,T)is weakly compatible andTu=Bu, we have

BTu=T Bu. i.e.,Bz=T z.

Now we prove thatBz=z. IfBz6=z, then

d(Bz,z) =d(z,Bz) =d(Az,Bz)

≤ψ(max{

d(Sz,Az)d(T z,Bz)

d(Sz,T z) +d(Sz,Bz) +d(T z,Az),d(Sz,T z)})

=ψ(max{0,d(z,Bz)}) =ψ(d(z,Bz))<d(z,Bz),

a contradiction. Hence,Bz=z. ThereforeBz=T z=z.

HenceAz=Bz=Sz=T z=z.

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Ifz0is also a common fixed point ofA,B,SandT withz6=z0,then

d(z,z0) =d(Az,Bz0)≤ψ(max{

d(Sz,Az)d(T z0,Bz0)

d(Sz,T z0) +d(Sz,Bz0) +d(T z0,Az),d(Sz,T z 0₎_}₎

=ψ(max{0,d(z,z0)}) =ψ(d(z,z0))<d(z,z0),

a contradiction. Therefore,z=z0.

Hence,zis the unique common fixed point ofA,B,SandT. The proof of (ii) is similar to (i) and hence is omitted.

Lemma 2.2. Let A,B,S and T be selfmaps of a metric space (X,d) and satisfy (1) and the inequality (3). Then for anyx₀∈X, the sequence{y_n}defined by (2) is Cauchy inX.

Proof.Letx₀∈X and let{y_n}be a sequence defined by (2). Assume thaty_n=y_n₊₁for somen.

Case(i): neven.

We writen=2m,m∈_N. Now we consider

d(y_n₊₁,y_n₊₂) =d(y₂_m₊₁,y₂_m₊₂)

=d(y₂_m+2,y2m+1)

=d(Ax2m+2,Bx2m+1)

≤ψ(max{ d(Sx2m+2,Ax2m+2)d(T x2m+1,Bx2m+1)

d(Sx₂_m₊₂,T x₂_m₊₁) +d(Sx₂_m₊₂,Bx₂_m₊₁) +d(T x₂_m₊₁,Ax₂_m₊₂),

d(Sx₂_m+2,T x2m+1)

=ψ(max{ d(y2m+1,y2m+2)d(y2m,y2m+1)

d(y₂_m₊₁,y₂_m) +d(y₂_m₊₁,y₂_m₊₁) +d(y₂_m,y₂_m₊₂),

d(y₂_m+1,y2m)})

≤ψ(max{d(y2m+1,y2m+2)d(y2m,y2m+1)

d(y₂_m₊₁,y₂_m₊₂) ,d(y2m+1,y2m)})

=ψ(d(y2m+1,y2m)) =ψ(0) =0.

Therefore,d(y₂_m₊₁,y₂_m₊₂) =0 which implies thaty₂_m₊₂=y₂_m₊₁=y₂_m. In general, we havey₂_m+k=y2mfork=0,1,2, . . . .

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We writen=2m+1 for somem∈_N. We consider

d(y_n₊₁,y_n₊₂) =d(y₂_m₊₂,y₂_m₊₃)

=d(Ax₂_m₊₂,Bx₂_m₊₃)

d(Sx2m+2,T x2m+3) +d(Sx2m+2,Bx2m+3) +d(T x2m+3,Ax2m+2) ,

d(Sx₂_m₊₂,T x₂_m₊₃)})

=ψ(max{ d(y2m+1,y2m+2)d(y2m+2,y2m+3)

d(y₂_m₊₁,y₂_m₊₂) +d(y₂_m₊₁,y₂_m₊₃) +d(y₂_m₊₂,y₂_m₊₂),

d(y2m+1,y2m+2)})

≤ψ(max{d(y2m+1,y2m+2)d(y2m+2,y2m+3)

d(y₂_m₊₂,y₂_m₊₃) ,d(y2m+1,y2m+2)})

=ψ(d(y2m+1,y2m+2)) =ψ(0) =0.

Therefore,d(y_n+2,yn+3) =0 implies thaty2m+3=y2m+2=y2m+1.

In general, we havey₂_m₊_k=y₂_m₊₁fork=1,2,3, . . . .

From Case (i) and Case (ii), we havey_n+k=ynfork=0,1,2, . . . .

Hence,{y_n₊_k}is a constant sequence and hence{y_n}is Cauchy.

Now we assume thatyn6=yn+1, for alln∈N. Ifnis odd, thenn=2m+1 for somem∈N.

We now consider

d(y_n,y_n₊₁) =d(y₂_m₊₁,y₂_m₊₂)

=d(y₂_m₊₂,y₂_m₊₁)

=d(Ax₂_m+2,Bx2m+1)

d(Sx₂_m₊₂,T x₂_m₊₁) +d(Sx₂_m₊₂,Bx₂_m₊₁) +d(T x₂_m₊₁,Ax₂_m₊₂),

d(Sx₂_m+2,T x2m+1)})

=ψ(max{ d(y2m+1,y2m+2)d(y2m,y2m+1)

d(y2m+1,y2m) +d(y2m+1,y2m+1) +d(y2m,y2m+2) ,

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≤ψ(max{d(y2m+1,y2m+2)d(y2m,y2m+1)

d(y₂_m₊₁,y₂_m₊₂) ,d(y2m+1,y2m)})

=ψ(d(y2m+1,y2m))<d(y2m+1,y2m)

Therefore,d(yn,yn+1)<d(yn−1,yn).

On the similar lines, ifnis even, it follows that

(4) d(y_n,y_n₊₁)≤ψ(d(yn−1,yn))<d(yn−1,yn).

Therefore,{d(y_n,y_n+1)}is a monotone decreasing sequence which bounded below by 0. So, there existsr≥0 such that lim

n→∞

d(y_n,y_n₊₁) =r.

Ifr>0, then from (4), we have

d(y_n,y_n₊₁)≤ψ(d(yn−1,yn)).

Lettingn→∞, we get r≤ψ(r)<r,

a contradiction. Therefore

(5) lim

n→∞

d(y_n,y_n₊₁) =0.

We now prove that{yn}is Cauchy.

It is sufficient to show that{y₂_n}is Cauchy inX.

Otherwise, there is anε>0 and there exist sequences{2mk},{2nk}with 2mk>2nk>k

such that

(6) d(y2mk,y2nk)≥ε andd(y2mk−2,y2nk)<ε.

Now, we prove that(i) lim

k→∞

d(y₂_m_k,y₂_n_k) =ε.

Sinceε ≤d(y2mk,y2nk)for allk,we have

ε≤lim inf

k→∞

d(y₂_m_k,y₂_n_k).

Now for each positive integerk, by the triangular inequality, we get

d(y2mk,y2nk)≤d(y2mk,y2mk−1) +d(y2mk−1,y2mk−2) +d(y2mk−2,y2nk). On taking limit superior ask→∞, from (5) and (6), we have

lim sup

k→∞

(8)

Hence, lim

k→∞

d(y₂_m_k,y₂_n_k)exists and lim

k→∞

d(y₂_m_k,y₂_n_k) =ε.

In similar way, it is easy to see that

(ii) lim

k→∞

d(y2mk+1,y2nk) =ε; (iii) lim

k→∞

d(y2mk,y2nk−1) =εand(iv) lim

k→∞

d(y2nk−1,y2mk+1) =ε.

We now consider

d(y₂_n_k,y₂_m_k₊₁) =d(Ax₂_n_k,Bx₂_m_k₊₁)

≤ψ(max{

d(Sx₂_n_k,Ax₂_n_k)d(T x₂_m_k₊₁,Bx₂_m_k₊₁)

d(Sx₂_n_k,T x₂_m_k₊₁) +d(Sx₂_n_k,Bx₂_m_k₊₁) +d(T x₂_m_k₊₁,Ax₂_n_k),

d(Sx2nk,T x2mk+1)}) (7)

=ψ(max{ d(y2nk−1,y2nk)d(y2mk,y2mk+1)

d(y₂_n_k₋₁,y₂_m_k) +d(y₂_n_k₋₁,y₂_m_k+1) +d(y2mk,y2nk) ,

d(y₂_n_k₋₁,y₂_m_k)}).

On lettingk→∞in (7), we getε ≤ψ(max{0,ε}) =ψ(ε)<ε,

a contradiction.

Therefore,{y_n}is a Cauchy sequence inX. The following is the main result of this paper.

Theorem 2.3. Let A,B,S and T be selfmaps on a complete metric space (X,d) and satisfy (1) and the inequality (3). If the pairs (A,S)and(B,T)are weakly compatible and one of the range setsS(X),T(X),A(X)andB(X)is closed, then for anyx₀∈X, the sequence{y_n}defined by (2) is Cauchy inX and lim

n→∞

yn=z(say), z∈X and zis the unique common fixed point of A,B,SandT.

Proof.By Lemma 2.2, the sequence{y_n}is Cauchy inX. SinceX is complete, there existsz∈X such that lim

n→∞

yn=z.Thus,

(8) lim

n→∞

y₂_n= lim

n→∞

Ax₂_n= lim

n→∞

T x₂_n₊₁=z

and

(9) lim

n→∞

y₂_n₊₁= lim

n→∞

Bx₂_n₊₁= lim

n→∞

Sx₂_n₊₂=z.

We now consider the following four cases.

Case(i). S(X)is closed.

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Now we claim thatAu=z. Suppose thatAu6=z. We now consider

d(Au,Bx2n+1)≤ψ(max{_d₍_Su_,_{T x}d(Su,Au)d(T x2n+1,Bx2n+1)

2n+1)+d(Su,Bx2n+1)+d(T x2n+1,Au),d(Su,T x2n+1)}).

On lettingn→∞, using (8) and (9), we get d(Au,z)≤ψ(max{0,0}) =ψ(0) =0.

HenceAu=z. Therefore,

(10) Au=z=Su.

Since the pair(A,S)is weakly compatible andAu=Su, we have

ASu=SAu.i.e.,Az=Sz. Now, we prove thatAz=z. IfAz6=z, then

d(Az,Bx₂_n+1)≤ψ(max{_d₍_Sz_,_{T x}₂d_n₊₁(Sz₎₊,Az_d)₍d_Sz(T x_,_Bx2₂n_n+1₊₁,Bx₎₊2_dn₍+1_{T x})₂_n₊₁_,_Az₎,d(Sz,T x2n+1)}). On lettingn→∞, using (8) and (9), we get

d(Az,z)≤ψ(max{0,d(Az,z)}) =ψ(d(Az,z))<d(Az,z),

a contradiction. Hence,Az=z. Therefore,Az=Sz=z.

Hence,zis a common fixed point ofAandS.

By Lemma 2.1, we get thatzis a unique common fixed point ofA,B,SandT.

Case(ii). T(X)is closed.

In this casez∈T(X)and there existsu∈X such thatz=Tu. Now we claim thatBu=z. Suppose thatBu6=z.

We now consider

d(Ax₂_n₊₂,Bu)≤ψ(max{_d₍_Sx d(Sx2n+2,Ax2n+2)d(Tu,Bu)

2n+2,Tu)+d(Sx2n+2,Bu)+d(Tu,Ax2n+2),d(Sx2n+2,Tu)}).

On lettingn→∞, using (8) and (9), we get

d(z,Bu)≤ψ(max{0,0}) =ψ(0) =0 and henceBu=z.

Therefore,

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Since the pair(B,T)is weakly compatible andBu=Tu, we have

BTu=T Bu.i.e.,Bz=T z.

Now, we prove thatBz=z. IfBz6=z, then

d(Ax₂_n₊₂,Bz)≤ψ(max{_d₍_Sx d(Sx2n+2,Ax2n+2)d(T z,Bz)

2n+2,T z)+d(Sx2n+2,Bz)+d(T z,Ax2n+2),d(Sx2n+2,T z)}).

d(z,Bz)≤ψ(max{0,d(z,Bz)}) =ψ(d(z,Bz))<d(z,Bz),

a contradiction. Hence,Bz=zand thatBz=T z=z. Therefore,zis a common fixed point ofBandT.

Hence, by Lemma 2.1, we get thatzis the unique common fixed point ofA,B,SandT.

Case(iii). A(X)is closed.

Sincez∈A(X)⊆T(X), there existsu∈X such thatz=Tu. Now we show thatBu=z.

IfBu6=z, then we consider

d(Ax₂_n+2,Bu)≤ψ(max{_d₍_Sx₂_n₊₂d_,_Tu(Sx₎₊2n+2_d₍_Sx,Ax₂_n2₊₂n+2_,_Bu)d(₎₊Tu_d,Bu₍_Tu)_,_Ax₂_n₊₂₎,d(Sx2n+2,Tu)}). On lettingn→∞, using (8) and (9), we get

d(z,Bu)≤ψ(max{0,0}) =ψ(0) =0 and henceBu=z.

Therefore Bu=z=Tu. Thus (11) holds. Now by Case (ii), the conclusion of the theorem follows.

Case(iv). B(X)is closed.

Sincez∈B(X)⊆S(X), there existsu∈X such thatz=Su. Now we show thatAu=z.

IfAu6=z, then we consider

d(Au,Bx₂_n₊₁)≤ψ(max{_d₍_Su_,_{T x}d(Su,Au)d(T x2n+1,Bx2n+1)

d(Au,z)≤ψ(max{0,0}) =ψ(0) =0 and henceAu=z.

ThereforeAu=z=Su. Thus (10) holds.

Now by Case(i), the conclusion of the theorem follows.

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(S(X),d),(T(X),d),(A(X),d) (or) (B(X),d) is complete, then for any x₀∈X, the sequence

{y_n} defined by (2) is Cauchy in X and lim

n→∞

y_n=z(say), z∈X andz is the unique common fixed point ofA,B,SandT

Proof.By Lemma 2.2, the sequence{yn}is Cauchy inX.

SinceS(X)is complete, there existsz∈S(X)such that lim

n→∞

y_n=z.Thus,

(12) lim

n→∞

y2n=_n→lim

∞

Ax2n=_n→lim

∞

T x2n+1=z

and

(13) lim

n→∞

y2n+1= lim

n→∞

Bx2n+1= lim

n→∞

Sx2n+2=z.

Sincez∈S(X), there existsu∈X such thatz=Su. We now prove thatAu=z. IfAu6=z, then

d(Au,Bx₂_n₊₁)≤ψ(max{_d₍_Su_,_{T x}d(Su,Au)d(T x2n+1,Bx2n+1)

d(Au,z)≤ψ(max{0,0}) =ψ(0) =0 and henceAu=z.

Therefore,Au=z=Su.

Since the pair(A,S)is weakly compatible andAu=Su, we have

ASu=SAu.i.e.,Az=Sz.

Now, we prove thatAz=z. If suppose thatAz6=z, then

d(Az,Bx2n+1)≤ψ(max{_d₍_Sz_,_{T x}d(Sz,Az)d(T x2n+1,Bx2n+1)

2n+1)+d(Sz,Bx2n+1)+d(T x2n+1,Az),d(Sz,T x2n+1)}).

d(Az,z)≤ψ(max{0,d(Az,z)}) =ψ(d(Az,z))<d(Az,z), a contradiction. Hence,Az=z. ThereforeAz=Sz=z. Thus,zis a common fixed point ofAandS.

By Lemma 2.1, we getzis the unique common fixed point ofA,B,SandT.

In a similar way, it is easy to see thatz is the unique common fixed point of A,B,S and T

when eitherT(X)orA(X)orB(X)is complete.

(12)

(i) (A,S) is reciprocally continuous and compatible pair of maps, and (B,T) is a pair of weakly compatible maps (or)

(ii) (B,T) is reciprocally continuous and compatible pair of maps, and (A,S) is a pair of weakly compatible maps.

ThenA,B,SandT have a unique common fixed point.

Proof.By Lemma 2.2, for eachx0∈X, the sequence{yn}defined by (2) is Cauchy inX.

SinceX is complete, then there existsz∈X such that lim

n→∞

y_n=z.

Consequently, the subsequences{y2n}and{y2n+1}are also converges toz∈X, we have

(14) lim

n→∞

y2n=_n→lim

∞

Ax2n=_n→lim

∞

T x2n+1=z,and

(15) lim

n→∞

y2n+1= lim

n→∞

Bx2n+1= lim

n→∞

Sx2n+2=z.

First, we assume that(i)holds.

Since(A,S)is reciprocal continuous, it follows that lim

n→∞

ASx₂_n+2=Azand lim

n→∞

SAx₂_n+2=Sz.

Since(A,S)is compatible, we have lim

n→∞

d(ASx₂_n+2,SAx2n+2) =0 which implies that lim

n→∞

d(Az,Sz) =0 implies thatAz=Sz. SinceA(X)⊆T(X), there existsu∈X such thatAz=Tu. Therefore,Az=Sz=Tu.

Now, we prove thatAz=Bu.Suppose thatAz6=Bu. We now consider

d(Az,Bu)≤ψ(max{ d(Sz,Az)d(Tu,Bu)

d(Sz,Tu) +d(Sz,Bu) +d(Tu,Az),d(Sz,Tu)})

=ψ(max{0,0}) =ψ(0) =0.

This implies thatAz=Bu=Sz=Tu.

Since every compatible pair is weakly compatible, we have (A,S) is weakly compatible and

Az=Sz, we haveASz=SAz.i.e,AAz=SAz.

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Now, we consider

d(AAz,Az) =d(AAz,Bu)≤ψ(max{ d(SAz,AAz)d(Tu,Bu)

d(SAz,Tu) +d(SAz,Bu) +d(Tu,AAz),d(SAz,Tu)})

=ψ(max{0,d(AAz,Az)})≤ψ(d(AAz,Az))<d(AAz,Az),

a contradiction.

ThereforeAAz=Az.Hence,AAz=SAz=Az,so thatAzis a common fixed point ofAandS. Since(B,T)is weakly compatible andBu=Tu, we haveBTu=T Bu.

ThereforeBAz=TAz.

We now prove thatBAz=Az. Suppose thatBAz6=Az. We now consider

d(BAz,Az) =d(Az,BAz)≤ψ(max{ d(Sz,Az)d(TAz,BAz)

d(Sz,TAz) +d(Sz,BAz) +d(TAz,Az),d(Sz,TAz)})

=ψ(max{0,d(Az,BAz)}) =ψ(d(Az,BAz))<d(Az,BAz),

a contradiction.

Hence,BAz=Az.ThereforeBAz=TAz=Az. Hence,AAz=BAz=SAz=TAz=Az.

ThereforeAzis a common fixed point ofA,B,SandT. Now, we show thatAz=z. IfAz6=z, then

d(Az,Bx₂_n+1)≤ψ(max{

d(Sz,Az)d(T x₂_n+1,Bx2n+1)

d(Sz,T x₂_n₊₁) +d(Sz,Bx₂_n₊₁) +d(T x₂_n₊₁,Az),d(Sz,T x2n+1)})

d(Az,z)≤ψ(max{0,d(Az,z)}) =ψ(d(Az,z))<d(Az,z), a contradiction.

Hence,Az=z. ThereforeAz=Bz=Sz=T z=z. Hence,zis a common fixed point ofA,B,SandT.

In a similar way, under the assumption(ii), we obtain the existence of common fixed point ofA,B,SandT. Uniqueness of common fixed point follows from the inequality (3).

Theorem 2.6. LetA,B,SandT be selfmaps on a complete metric space(X,d)and satisfy (1) and the inequality (3). If either

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(ii) T is continuous,(B,T)compatible and(A,S)is a pair of weakly compatible maps, thenA,B,SandT have a unique common fixed point.

Proof.By Lemma 2.2, for eachx₀∈X, the sequence{y_n}defined by (2) is Cauchy inX. SinceX is complete, then there existsz∈X such that lim

n→∞

y_n=z.

Consequently, the subsequences{y₂_n}and{y₂_n₊₁}are also converges toz∈X, we have

(16) lim

n→∞

y₂_n= lim

n→∞

Ax₂_n= lim

n→∞

T x₂_n₊₁=z,

and

(17) lim

n→∞

y₂_n₊₁= lim

n→∞

Bx₂_n₊₁= lim

n→∞

Sx₂_n₊₂=z.

First, we assume that(i)holds.

Since(A,S)is compatible pair, we have lim

n→∞

d(SAx₂_n,ASx₂_n) =0, it follows that lim

n→∞

SAx2n=_n→lim

∞

ASx2n.

SinceSis continuous, we haveSz= lim

n→∞

SAx₂_n= lim

n→∞

ASx₂_n

Now, we prove thatSz=z. IfSz6=z, then consider

d(ASx₂_n₊₂,Bx₂_n₊₁)≤ψ(max{_d₍_SSx d(SSx2n+2,ASx2n+2)d(T x2n+1,Bx2n+1)

2n+2,T x2n+1)+d(SSx2n+2,Bx2n+1)+d(T x2n+1,ASx2n+2),d(SSx2n+2,T x2n+1)})

d(Sz,z)≤ψ(max{_d₍_Sz_,_zd₎₊(Sz_d,₍Sz_Sz)_,d_z(₎₊z,z_d)₍_z_,_Sz₎,d(Sz,z)}) =ψ(d(Sz,z))<d(Sz,z),

a contradiction. Hence,Sz=z.

We now prove thatAz=z. If possible, suppose thatAz6=z. Now we consider

d(Az,Bx₂_n+1)≤ψ(max{_d₍_Sz_,_{T x}₂d_n₊₁(Sz₎₊,Az_d)₍d_Sz(T x_,_Bx2₂n_n+1₊₁,Bx₎₊2_dn₍+1_{T x})₂_n₊₁_,_Az₎,d(Sz,T x2n+1)}) On taking limits asn→∞, using (16) and (17), we get

d(Az,z)≤ψ(max{_d₍_z_,_zd₎₊(z_d,Az₍_z_,)_zd₎₊(z,_dz₍)_z_,_Az₎,d(z,z)}) =ψ(max{0,0}) =ψ(0) =0.

Therefored(Az,z)≤0 which implies thatAz=z. HenceAz=Sz=z.

Thereforezis a common fixed point ofAandS.

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In a similar way, under the assumption (ii), we can obtain the existence of common fixed point ofA,B,SandT. Uniqueness of common fixed point follows from the inequality (3).

3. Corollaries and examples

In this section, we draw some corollaries from the main results of Section 2 and provide examples in support of our results.

The following is an example in support of Theorem 2.3.

Example 1.LetX= [0,1]with usual metric. We define selfmapsA,B,S, T onX by

A(x) =  

 x2

2 if 0≤x< 1 2

0 if 1₂≤x≤1

,B(x) =  

 x2

4 if 0≤x< 1 2

0 if 1₂ ≤x≤1

S(x) =  



x2 if 0≤x< 1₂ 1 if 1₂ ≤x≤1

andT(x) =  

 x2

2 if 0≤x< 1 2

0 if 1₂≤x≤1. We defineψ :_R+→R+ byψ(t) =₂t, t≥0.Then clearlyψ ∈Ψ.

HereA(X) = [0,1₈),B(X) = [0,₁₆1),

S(X) = [0,1₄)∪ {1}andT(X) = [0,1₈). So thatA(X)⊆T(X)andB(X)⊆S(X). We now verify the inequality (3).

Without loss of generality assume thatx≥y.

Case(i): x,y∈[0,1₂).

d(Ax,By) =|x₂2 −y₄2|;d(Sx,Ty) =|x2−y₂2|. We consider

d(Ax,By) =1

2|x 2₋y2

2|= 1

2d(Sx,Ty)

=ψ(d(Sx,Ty))

≤ψ(max{ d(Sx,Ax)d(Ty,By)

d(Sx,Ty) +d(Sx,By) +d(Ty,Ax),d(Sx,Ty)}).

Case(ii): x,y∈[1₂,1].

d(Ax,By) =0 and trivially holds the inequality (3) in this case.

Case(iii): x∈[0,1₂),y∈[1₂,1].

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We have

d(Ax,By) =x

2

2 = 1

2d(Sx,Ty) =ψ(d(Sx,Ty))

Case(iv): x∈[1₂,1],y∈[0,1₂).

d(Ax,By) =y₄2;d(Sx,Ty) = (1−y₂2). We now consider

d(Ax,By) =y

2

4 ≤ 1 2(1−

y2

2) = 1

2d(Sx,Ty)

=ψ(d(Sx,Ty))

From the above four cases,A,B,SandT satisfy the inequality (3).

ThereforeA,B,SandT satisfy all the hypotheses of Theorem 2.3 and 0 is the unique common fixed point ofA,B,SandT.

Corollary 3.1. Let{A_n}∞

n=1,SandT be selfmaps on a complete metric space(X,d)satisfying

A₁⊆S(X)andA₁⊆T(X). Assume that there existsψ ∈Ψsuch that

(18) d(A1x,Ajy)≤       

     

ψ(max{_d₍_Sx_,_Tyd(₎₊Sx_d,A₍1_Sxx)_,_Ad(Ty,Ajy)

jy)+d(Ty,A1x),d(Sx,Ty)}),

if d(Sx,Ty) +d(Sx,A_jy) +d(Ty,A₁x)6=0 0, if d(Sx,Ty) +d(Sx,A_jy) +d(Ty,A₁x) =0,

for all x,y∈X and j=1,2,3, . . . .If the pairs (A₁,S) and(A₁,T)are weakly compatible and one of the range sets A₁(X),S(X) and T(X) is closed, then {A_n}∞

n=1,S and T have a unique common fixed point inX.

Proof.Under the assumptions onA₁,SandT, the existence of common fixed pointzofA₁,SandT

follows by choosingA=B=A₁in Theorem 2.3. ThereforeA₁z=Sz=T z=z.

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We now consider

d(z,A_jz) =d(A₁z,A_jz)

≤ψ(max{

d(Sz,A₁z)d(T z,A_jz)

d(Sz,T z) +d(Sz,A_jz) +d(T z,A₁z),d(Sz,T z)})

=ψ(max{0,0}) =ψ(0) =0.

Therefored(z,A_jz)≤0 which implies thatA_jz=zfor j=1,2,3, . . .and uniqueness of common fixed point follows from the inequality (18).

Hence,{A_n}∞

n=1,SandT have a unique common fixed point inX. Corollary 3.2.Let{A_n}∞

n=1,SandT be selfmaps on a metric space(X,d)satisfy the conditionsA₁⊆S(X),A₁⊆T(X)and (18). If the pairs(A₁,S)and(A₁,T)are weakly compatible and either (A₁(X),d),(S(X),d) or (T(X),d) is complete, then {A_n}∞

n=1,S and T have a unique common fixed point inX.

Proof.Under the assumptions onA1,SandT, the existence of common fixed pointzofA1,SandT follows by choosingA=B=A₁in Theorem 2.4.

ThereforeA1z=Sz=T z=z. Now, let j∈_Nwith j6=1. We now consider

d(z,A_jz) =d(A₁z,A_jz)

≤ψ(max{

d(Sz,A₁z)d(T z,A_jz)

d(Sz,T z) +d(Sz,A_jz) +d(T z,A₁z),d(Sz,T z)})

=ψ(max{0,0}) =ψ(0) =0.

Therefored(z,Ajz)≤0 which implies thatAjz=zfor j=1,2,3, . . .and uniqueness of common

fixed point follows from the inequality (18).

Hence,{An}∞_n₌₁,SandT have a unique common fixed point inX.

The following is an example in support of Theorem 2.5. In this example we show the importance of rational expression in the inequality (3).

Example 2.LetX= [0,42₂₇]∪[46₂₇,2]with usual metric. We define selfmapsA,B,S, T onX by

A(x) =  



2

3 ifx∈[0, 42 27]∪[

46 27,2) 1

2 ifx=2

,B(x) =  



2

3 ifx∈[0, 42 27]∪[

46 27,2) 1

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S(x) =          2

3 if 0≤x< 2 3 4

3−x if 2

3≤x≤1

0 ifx∈(1,42₂₇]∪[46₂₇,2]

andT(x) =          1

3 ifx=0 1−x

2 ifx∈(0, 42 27]∪[

46 27,2) 4

3 ifx=2

We defineψ :_R+→R+ byψ(t) =3₄t, t≥0.Then clearlyψ∈Ψ.HereA(X) ={1₂,2₃}, B(X) ={1₃,2₃},S(X) = [1₃,2₃]∪ {0}andT(X) = (0,₂₇4]∪[₂₇5,1)∪ {4₃}so thatA(X)⊆T(X)and

B(X)⊆S(X).

We now verify the inequality (3).

Case(1): x=y=0.

d(Ax,By) =0 and trivially holds the inequality (3). Case (2): x=y=2.

d(Ax,By) =1₆;d(Sx,Ty) =4₃. We consider

d(Ax,By) = 1

6≤ 3 4(

4

3) =ψ(d(Sx,Ty))

Case(3): x,y∈(0,2₃).

d(Ax,By) =0 and trivially holds the inequality (3).

Case(4): x,y∈[2₃,1].

Case(5): x,y∈(1,42₂₇]∪[46₂₇,2).

Case(6):x=0,y=2.

d(Ax,By) =1₃;d(Sx,Ty) =2₃.We now consider

d(Ax,By) = 1

3≤ 3 4(

2

3) =ψ(d(Sx,Ty))

Case(7):x=0,y∈(0,2₃).

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Case(9): x=0,y∈(1,42₂₇]∪[46₂₇,2).

Case(10): x=2,y=0.

d(Ax,By) =1₆;d(Sx,Ty) =1₃.We have

d(Ax,By) = 1

6≤ 3 4(

1

3) =ψ(d(Sx,Ty))

Case(11): x=2,y∈(0,2₃).

d(Ax,By) =1₆;d(Sx,Ty) =1−₂y. We consider

d(Ax,By) = 1

6≤ 3 4(1−

y

2) =ψ(d(Sx,Ty))

Case(12): x=2,y∈[2₃,1].

d(Ax,By) =1₆;d(Sx,Ty) =1−₂y.We have

d(Ax,By) = 1

6≤ 3 4(1−

y

2) =ψ(d(Sx,Ty))

Case(13): x=2,y∈(1,42₂₇]∪[46₂₇,2).

d(Ax,By) =1₆;d(Sx,Ty) =|1−₂y|.

d(Sx,Ax) = 1₂;d(Ty,By) =|1₃−y₂|.

d(Sx,By) = 2₃;d(Ty,Ax) =|1₂−y₂|.

Subcase(i).x=2 andy∈(1,42₂₇].

d(Ax,By) = 1

6≤ 3 4(1−

y

2) =ψ(d(Sx,Ty))

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We now consider

d(Ax,By) = 1

6≤ 3 4(

3y−2

14 ) =ψ(

d(Sx,Ax)d(Ty,By)

d(Sx,Ty) +d(Sx,By) +d(Ty,Ax))

Case(14): x∈(0,2₃),y=0.

Case(15): x∈(0,2₃),y=2.

d(Ax,By) =1₃;d(Sx,Ty) =2₃. We consider

d(Ax,By) = 1

3≤ 3 4(

2

3) =ψ(d(Sx,Ty))

Case(16):x∈(0,2₃),y∈[2₃,1].

Case(17):x∈(0,2₃),y∈(1,42₂₇]∪[46₂₇,2).

Case(18): x∈[2₃,1],y=0.

Case(19): x∈[2₃,1],y=2.

d(Ax,By) =1₃;d(Sx,Ty) =x. We have

d(Ax,By) = 1

3 ≤ 3

4(x) =ψ(d(Sx,Ty))

Case(20): x∈[2₃,1],y∈(0,2₃).

Case(21): x∈[2₃,1],y∈(1,42₂₇]∪[46₂₇,2).

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Case(23): x∈(1,42₂₇]∪[46₂₇,2),y=2.

d(Ax,By) =1₃;d(Sx,Ty) =4₃. We now consider

d(Ax,By) = 1

3≤ 3 4(

4

3) =ψ(d(Sx,Ty))

Case(24): x∈(1,42₂₇]∪[46₂₇,2),y∈(0,2₃).

d(Ax,By) =0 and trivially holds the inequality in this case.

Case(25): x∈(1,42₂₇]∪[46₂₇,2),y∈[2₃,1].

d(Ax,By) =0 and trivially holds the inequality (3) in this case. From the above all cases,A,B,SandT satisfy the inequality (3).

ThereforeA,B,SandT satisfy all the hypotheses of Theorem 2.5 and 2₃ is the unique common fixed point ofA,B,SandT.

Observation: Subcase (ii) of Case (13) indicates the importance of the rational expression in the inequality (3), since in the absence of the rational expression, the inequality (3) fails to hold, ford(Ax,By) = 1₆ψ(|1−y₂|)for anyψ ∈Ψ.

Example 3.LetX= [0,1]with usual metric. We define selfmapsA,B,S, T onX by

A(x) =  

 x2

2 if 0≤x< 1 2

0 if 1₂≤x≤1,

,B(x) =  

 x2

4 if 0≤x< 1 2

0 if 1₂ ≤x≤1,

S(x) =x2if 0≤x≤1 andT(x) =  

 x2

2 if 0≤x< 1 2

0 if 1₂ ≤x≤1. We defineψ :_R+→R+ byψ(t) =₂t, t≥0.

Then clearlyψ∈Ψ.

HereA(X) = [0,1₈),B(X) = [0,₁₆1),S(X) = [0,1]andT(X) = [0,1₈). ClearlyA(X)⊆T(X)andB(X)⊆S(X).

Without loss of generality assume thatx≥y.

Case(i): x,y∈[0,1₂).

d(Ax,By) =|x₂2 −y₄2|=x₂2 −y₄2;

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We consider

d(Ax,By) =1

2|x 2₋y2

2|=ψ(d(Sx,Ty))

Case(ii): x,y∈[1₂,1].

Case(iii): x∈[0,1₂),y∈[1₂,1].

d(Ax,By) =x₂2;d(Sx,Ty) =x2. We have

d(Ax,By) =x

2

2 = 1

2d(Sx,Ty) =ψ(d(Sx,Ty))

Case(iv): x∈[1₂,1],y∈[0,1₂).

d(Ax,By) =y₄2;d(Sx,Ty) = (x2−y₂2). We now consider

d(Ax,By) = y

2

4 ≤ 1 2(x

2₋y2

2) =ψ(d(Sx,Ty))

From the above four cases,A,B,SandT satisfy the inequality (3).

ThereforeA,B,SandT satisfy all the hypotheses of Theorem 2.6 and 0 is the unique common fixed point ofA,B,SandT.

Corollary 3.3.LetT,f,g:X→X satisfying

d(T x,f y)≤ψ(max{ d(gx,T x)d(gy,f y)

d(gx,gy) +d(gx,f y) +d(gy,T x),d(gx,gy)})

for allx,y∈X,T(X)∪f(X)⊆g(X)and(g(X),d)is complete. ThenT,f andghave a

coincidence point in X. If the pairs (T,g) and (f,g)are weakly compatible, then T,f andg

have a unique common fixed point inX.

Proof. By choosing A=T,B = f,S= T = g in Theorem 2.4, the conclusion of corollary follows.

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An open problem: Is the conclusion of Theorem 2.6 valid if we replace ‘continuity ofS’ by ‘continuity ofA’? Similarly, is the conclusion of Theorem 2.6 valid if we replace ‘continuity of

T’ by ‘continuity ofB’? Conflict of Interests

The authors declare that there is no conflict of interests. REFERENCES

[1] S. Chandok, Common fixed point theorems for generalized contraction mappings, Thai J. Math., (In press). [2] B. K. Dass and S. Gupta, An extension of Banach contraction principle through rational expressions, Indian

J. Pure and Appl. Math., 6(1975), 1455-1458.

[3] G. Jungck, Compatible mappings and common fixed points, Internat. J. Math. and Math. Sci., 9(1986), 771-779.

[4] G. Jungck and B. E. Rhoades, Fixed points of set-valuaed functions without continuity, Indian J. Pure and Appl. Math., 29(3)(1998), 227-238.