# Some Common Fixed Point Theorems in Generalized Metric Spaces

Authors: Manoj Kumar; Pankaj Kumar; Sanjay Kumar
Journal: Journal of Mathematics
 (2013)

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## Abstract

First we prove common fixed point theorems for weakly compatible maps which generalize the results of Chen (2012). Secondly, we prove common fixed point theorems using property E.A. along with weakly compatible maps. At the end, we prove common fixed point theorems using common limit range property (CLR property) along with weakly compatible maps.


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## Body

## 1. Introduction

In 2000, Branciari [1] introduced the notion of generalized metric space as follows.Definition 1.
LetX be a nonempty set and let  d:X×X→[0,∞) be a mapping such that for all x,y∈X and for all distinct points  u,v∈X  each of them different from x and y, one has the following:(i)
d(x,y)=0 if and only if x=y,(ii)
d(x,y)=d(y,x),(iii)
d(x,y)≤d(x,u)+d(u,v)+d(v,y) (rectangular inequality).
Then(X,d) is called a generalized metric space (or shortly g.m.s.).Definition 2.
Let(X,d) be a g.m.s. Let {xn} be a sequence in X and  x∈X. We say that {xn} is (i)
g.m.s. convergent tox if and only if  d(xn,x)→0 as  n→∞,(ii)
g.m.s. Cauchy sequence if and only if, for eachε>0, there exists n0∈ℕ such that, d(xm,xn)<ε for all n>m>n0,(iii)
complete g.m.s. if every g.m.s. Cauchy sequence is g.m.s. convergent inX.
In 1986, Jungck [2] introduced the notion of compatible maps as follows.
Let(X,d) be a metric space. Let S,F:X→X be two single-valued functions. We say that S and F are compatible if  limn→∞d(SFxn,FSxn)=0, whenever {xn} is a sequence in X such that limn→∞d(Fxn,Sxn)=0.
In 1998, Jungck and Rhoades [3] introduced the notion of weakly compatible maps as follows.
Two maps are said to be weakly compatible if they commute at the coincidence point. Note that compatible maps are weakly compatible but the converse is not true in general.
In 2002, Aamri and El Moutawakil [4] introduced the notion of property E.A. as follows.
LetA and B be two self-mappings of a metric space (X,d). We say that A and B satisfy the property E.A. if there exists a sequence {xn} such that limn→∞Axn=limn→∞Bxn=t for some t in X.
In 2011, Sintunavarat and Kumam [5] introduced the notion of (CLRg) property as follows.
Let(X,d) be a metric space. Two mappings f:X→X and g:X→X are said to satisfy (CLRg) property if there exists sequences {xn} in X such that
(1)limn→∞f(xn)=limn→∞g(xn)=g(t)forsometinX.
Similarly, we can state (CLRF) and (CLRG) properties if the mapping  g:X→X  has been replaced by the mapping F:X→X and  G:X→X, respectively.
In 2012, Chen [6] introduced the notion of 𝒲 function and 𝒮 function as follows.Definition 3.
A functionφ:ℝ+→ℝ+ is said to be a  𝒲  function if the function ϕ satisfies the following conditions:(ϕ1) ϕ(t)<t for all t>0 and ϕ(0)=0,(ϕ2) limtn→tinfϕ(tn)<t for all t>0.Definition 4.
A functionφ:ℝ+→ℝ+ is said to be a 𝒮  function if the function ϕ satisfies the following conditions:(ϕ1) ϕ is a strictly increasing, continuous function in each coordinate,(ϕ2) for all t>0, ϕ(t,t,t)<t,  ϕ(t,0,0)<t,  ϕ(0,t,0)<t, and ϕ(0,0,t)<t.
Before proving our results, we need the following lemma.Lemma 5 (see [6]).
Letϕ:ℝ+→ℝ+ be a 𝒲 function. Then limn→∞ϕn(t)=0for all t>0, where ϕn(t) denotes the nth iteration of ϕ.


## 2. Weakly Compatible Maps

In 2012, Chen [6] proved the following fixed point theorems for compatible maps.Theorem 6.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+→ℝ+ be a 𝒲 function. Let S,  T,  F,  and  G  be self-maps on X satisfying the following:
(2)T(X)⊆F(X),S(X)⊆G(X),(3)d(Sx,Ty)≤ϕ(max{d(Fx,Gy),d(Fx,Sx),d(Gy,Ty)})∀x,y∈X,(4)pairs{S,F}and{T,G}arecompatible.
If F or G is continuous, then S, T,  F, and G have a unique common fixed point in X.Theorem 7.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+3→ℝ+  be a 𝒮 function. Let  S,  T, F, and G be self-maps on X satisfying (2), (4), and the following:
(5)d(Sx,Ty)≤ϕ({d(Fx,Gy),d(Fx,Sx),d(Gy,Ty)})∀x,y∈X.
If one of S(X), T(X), F(X), and G(X), is a complete subset of X then S, T, F, and G have a unique common fixed point in X.Now, we prove our results for weakly compatible maps.Theorem 8.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+→ℝ+ be a 𝒲 function. Let S, T,  F, and G be self-maps on X satisfying (2), (3) and the following:
(6)pairs{S,F}and{T,G}areweaklycompatible.
If one of S(X), T(X),  F(X), and G(X) is a complete subset of X, then S,  T,  F, and G have a unique common fixed point in X.Proof.
Letx0∈X. Define the sequence {xn} recursively as follows:
(7)Gx2n+1=Sx2n=z2n,Fx2n+2=Tx2n+1=z2n+1.
By Theorem 6 [6], {zn} is g.m.s. Cauchy sequence in X. Now suppose that F(X) is complete subset of X, then the subsequence z2n+1=Fx2n+2 must get a limit in F(X). Call it to be u and v∈F-1(u). Then F(v)=u. As {zn} is a Cauchy sequence containing a convergent subsequence {z2n+1}, therefore the sequence {zn} also converges implying, thereby, the convergence of {z2n}  being a subsequence of the convergent sequence {zn}.
Takingx=v and y=x2n+1 in (3), we get
(8)d(Sv,Tx2n+1)≤ϕ(max{d(Fv,Gx2n+1),d(Fv,Sv),d(Gx2n+1,Tx2n+1)}).
Letting n→∞, we have
(9)d(Sv,u)≤ϕ(max{d(u,u),d(u,Sv),d(u,u)})=ϕ(d(u,Sv))<d(Sv,u),acontradiction.
Therefore, Sv=u=Fv implies that the pair (S,F) has a point of coincidence.
AsS(X)⊆G(X), Sv=u implies that u∈G(X). Letting w∈G-1(u), then Gw=u.
Now, from (3), we have
(10)d(z2n,Tw)=d(Sx2n,Tw)≤ϕ(max{d(Fx2n,Gw),d(Fx2n,Sx2n),d(Gw,Tw)Fx2n}).
Letting n→∞, we have
(11)d(u,Tw)≤ϕ(max{d(u,u),d(u,u),d(u,Tw)})=ϕ(d(u,Tw))<d(u,Tw),acontradiction.
Therefore, u=Tw.
Thus, we have shown thatu=Sv=Fv=Tw=Gw, which amounts to say that both pairs have point of coincidence. If one assumes G(X) to be complete, then an analogous argument establishes this claim.
The remaining two cases pertain essentially to the previous cases. Indeed, ifS(X) is complete, then u∈S(X)⊆G(X), and if T(X) is complete, then u∈T(X)⊆F(X). Thus S and F,  T, and G have a point of coincidence.
Since the pairs(S,F) and (T,G) are weakly compatible at v and w, respectively, then
(12)Su=S(Fv)=F(Sv)=Fu,Tu=T(Gw)=G(Tw)=Gu.
If Su≠u, then
(13)d(Su,u)=d(Su,Tw)≤ϕ(max{d(Fu,Gw),d(Fu,Su),d(Gw,Tw)})=ϕ(max{d(Su,u),0,0})=ϕ(d(Su,u))<d(Su,u),acontradiction.
Therefore, Su=u.
Similarly, one can show thatTu=u.
Thusu is the common fixed point of S, T, F, and G.
Finally, we prove thatS, T, F, and G have a unique common fixed point. Let t be another common fixed point of S, T,  F, and G.
From (3), we have
(14)d(t,u)=d(St,Tu)≤ϕ(max{d(Ft,Gu),d(Ft,St),d(Gu,Tu)})=ϕ(max{d(t,u),d(t,t),d(u,u)})<d(t,u),acontradictionunlessd(t,u)=0,thatis,t=u.
Hence u is the unique common fixed point of S, T, F, and G in X.
We give the following example to illustrate Theorem8.Example 9.
LetX={a1,a2,a3,a4,a5}, where a1, a2, a3, a4, and  a5  are positive constants. We define d:X×X→[0,∞) by(1)
d(x,x)=0, for all x∈X,(2)
d(x,y)=d(y,x), for all x,y∈X,(3)
d(a1,a2)=3k,(4)
d(a1,a3)=d(a2,a3)=k,(5)
d(a1,a4)=d(a2,a4)=d(a3,a4)=2k,(6)
d(a1,a5)=d(a2,a5)=d(a3,a5)=d(a4,a5)=2k,
wherek>0 is a constant.
Ifϕ:ℝ+→ℝ+, ϕ(t)=(2/3)t, then ϕ is a 𝒲  function. We next define S,T,F,G:X→X by
(15)S(x)={a3,ifx≠a4,a5,ifx=a4,T(x)={a3,ifx≠a4,a1,ifx=a4,G(x)=I(x)= the identity mapping,
(16)F(x)={a3,ifx=a3,a1,ifx=a1,a2,a5,a2,ifx=a4.
Then all the conditions of Theorem 8 are satisfied and a3 is a unique common fixed point of S, T, F, and G.Theorem 10.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+3→ℝ+ be a 𝒮 function. Let S,  T, F, and G be self-maps on X satisfying (2), (5) and (6). If one of S(X), T(X), F(X), and G(X) is a complete subset of X, then S, T, F, and G have a unique common fixed point in X.Proof.
Given thatx0∈X, define the sequence {xn} recursively as follows:
(17)Gx2n+1=Sx2n=z2n,Fx2n+2=Tx2n+1=z2n+1.
Then due to Theorem 7 [6], {zn} is g.m.s. Cauchy sequence in X. Now, suppose that F(X) is complete subset of X; then the subsequence z2n+1=Fx2n+2 must get a limit in F(X). Call it to be u and v∈F-1(u). Then F(v)=u. As {zn} is a Cauchy sequence containing a convergent subsequence {z2n+1}, therefore the sequence {zn} also converges implying thereby the convergence of {z2n} being a subsequence of the convergent sequence {zn}. On taking x=v and y=x2n+1 in (5) one gets
(18)d(Sv,Tx2n+1)≤ϕ({d(Fv,Gx2n+1),d(Fv,Sv),d(Gx2n+1,Tx2n+1)}).
Letting n→∞, we have
(19)d(Sv,u)≤ϕ({d(u,u),d(u,Sv),d(u,u)})=ϕ(d(u,Sv))<d(Sv,u),acontradiction.
Therefore, Sv=u=Fv, which shows that the pair (S,F) has a point of coincidence.
AsS(X)⊆G(X), Sv=u implies that u∈G(X). Letting w∈G-1(u), then Gw=u.
From (5), we have
(20)d(z2n,Tw)=d(Sx2n,Tw)≤ϕ({d(Fx2n,Gw),d(Fx2n,Sx2n),d(Gw,Tw)Fx2n}).
Letting n→∞, we have
(21)d(u,Tw)≤ϕ({d(u,u),d(u,u),d(u,Tw)})=ϕ(d(u,Tw))<d(u,Tw),acontradiction.
Therefore, u=Tw. Thus, we have shown that u=Sv=Fv=Tw=Gw, which amounts to say that both pairs have point of coincidence. If one assumes G(X) to be complete, then an analogous argument establishes this claim.
The remaining two cases pertain essentially to the previous cases. Indeed ifS(X) is complete, then u∈S(X)⊆G(X), and if T(X) is complete, then u∈T(X)⊆F(X).
ThusS and F,  T, and G have a point of coincidence.
Since the pairs(S,F) and (T,G) are weakly compatible at v and w, respectively, then
(22)Su=S(Fv)=F(Sv)=Fu,Tu=T(Gw)=G(Tw)=Gu.
If Su≠u, then
(23)d(Su,u)=d(Su,Tw)≤ϕ({d(Fu,Gw),d(Fu,Su),d(Gw,Tw)})=ϕ({d(Su,u),0,0})=ϕ(d(Su,u))<d(Su,u),acontradiction.
Therefore, Su=u.
Similarly, one can show thatTu=u.
Thusu is the common fixed point of S, T, F, and G.
Finally, we prove thatS, T,  F, and  G  have a unique common fixed point. Let t be another common fixed point of S, T,  F, and G. Then using (5), we have
(24)d(t,u)=d(St,Tu)≤ϕ({d(Ft,Gu),d(Ft,St),d(Gu,Tu)})=ϕ({d(t,u),d(t,t),d(u,u)})<d(t,u),acontradictionunlessd(t,u)=0,thatis,t=u.
Hence u is the unique common fixed point of S, T,  F, and G in X.We give the following example to illustrate Theorem10.Example 11.
LetX={a1,a2,a3,a4,a5}, where a1, a2, a3,  a4, and a5  are positive constants. We define d:X×X→[0,∞) by(1)
d(x,x)=0, for all x∈X,(2)
d(x,y)=d(y,x), for all x,y∈X,(3)
d(a1,a2)=3k,(4)
d(a1,a3)=d(a2,a3)=k,(5)
d(a1,a4)=d(a2,a4)=d(a3,a4)=2k,(6)
d(a1,a5)=d(a2,a5)=d(a3,a5)=d(a4,a5)=2k,
wherek>0 is a constant.
Ifϕ:ℝ+3→ℝ+,   ϕ(t)=(3/5)·max{t1,t2,t3}, then ϕ is a 𝒮 function. We next define S, T, F,  G:X→X by
(25)S(x)={a3,ifx≠a4,a5,ifx=a4,T(x)={a3,ifx≠a4,a1,ifx=a4,G(x)=I(x)=theidentitymapping,F(x)={a3,ifx=a3,a1,ifx=a1,a2,a5,a2,ifx=a4.
Then all the conditions of Theorem 10 are satisfied and a3 is a unique common fixed point of S, T, F, and G.


## 3.  Property E.A.

Theorem 12.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+→ℝ+ be a 𝒲 function. Let S, T,  F, and  G be self-maps on X satisfying (2), (3), (6), and the following:
(26){S,F}or{T,G}satisfythepropertyE.A.
If the range of one of the maps S, T, F, or G is a complete subspace of X, then S, T, F, and G have a unique common fixed point in X.Proof.
If the pair{T,G} satisfies the property E.A., then there exists a sequence {xn} in X such that Txn→z and Gxn→z for some z∈X as n→∞. Since T(X)⊆F(X), there exists a sequence {yn} in X such that Txn=Fyn. Hence Fyn→z as n→∞. Also, since S(X)⊆G(X), there exists a sequence {yn′} in X such that Syn′=Gyn′. Hence Syn′→z as n→∞.
Suppose thatF(X) is a complete subspace of X. Then, z=Fu for some u∈X.
Subsequently, we haveSyn′→Fu, Txn→Fu,  Fyn→Fu as n→∞.
From (3), we have
(27)d(Su,Txn)≤ϕ(max{d(Fu,Gxn),d(Fu,Su),d(Gxn,Txn)}).
Letting n→∞, we have
(28)d(Su,Fu)≤ϕ(max{d(Fu,Fu),d(Fu,Su),d(Fu,Fu)})=ϕ(d(Fu,Su))<d(Fu,Su),
which implies that Fu=Su.
The weak compatibility ofF and S implies that FSu=SFu and then SSu=SFu=FSu=FFu.
On the other hand, sinceS(X)⊆G(X), there exists a v∈X such that Su=Gv.
Now, we show thatGv=Tv.
From (3), we have
(29)d(Su,Tv)≤ϕ(max{d(Fu,Gv),d(Fu,Su),d(Gv,Tv)})=ϕ(d(Gv,Tv))<d(Gv,Tv),impliesthatGv=Tv,thatis,Su=Fu=Gv=Tv.
The weak compatibility of T and G implies that TGv=GTv and GGv=GTv=TGv=TTv.
Let us show thatSu is the common fixed point of S,  T,  F,  and  G.
From (3), we have
(30)d(SSu,Tv)≤ϕ(max{d(FSu,Gv),d(FSu,SSu),d(Gv,Tv)})=ϕ(d(SSu,Tv))<d(Gv,Tv),impliesthatSSu=Tv=Su.
Therefore, Su=SSu=FSu and Su is the common fixed point of S and F.
Similarly, we can prove thatTv is the common fixed point of T and G.
SinceSu=Tv, we conclude that Su is the common fixed point of S,  T,  F,  and G.
The proof is similar whenG(X) is assumed to be complete subspace of X. The cases in which S(X) or T(X) is a complete subspace of X are similar to the cases in which G(X) or F(X), respectively, is complete since
(31)S(X)⊆G(X),T(X)⊆F(X).
If Su=Tu=Fu=Gu=u and Sv=Tv=Fv=Gv=v, then (3) gives
(32)d(Su,Tv)≤ϕ(max{d(Fu,Gv),d(Fu,Su),d(Gv,Tv)})<d(Fu,Gv),impliesthatu=v.
Therefore, u=v and the common fixed point is unique.Theorem 13.
Let(X,d) be Hausdorff and complete g.m.s. and let  ϕ:ℝ+3→ℝ+  be a 𝒮 function. Let S,  T, F, and G be self-maps on X satisfying (2), (5), (6), and (26). If the range of one of the maps S,  T, F, or G is a complete subspace of X, then S, T, F, and G have a unique common fixed point in X.Proof.
If the pair{T,G} satisfies the property E.A., then there exists a sequence {xn} in X such that Txn→z and Gxn→z for some z∈X as n→∞.
SinceT(X)⊆F(X), there exists a sequence {yn} in X such thatTxn=Fyn. Hence Fyn→z as n→∞. Also, since S(X)⊆G(X), there exists a sequence {yn′} in X such that Syn′=Gyn′. Hence Syn′→z as n→∞.
Suppose thatF(X) is a complete subspace of X. Then, z=Fu for some u∈X.
Subsequently, we haveSyn′→Fu, Txn→Fu, Fyn→Fu as n→∞.
From (5), we have
(33)d(Su,Txn)≤ϕ(d(Fu,Gxn),d(Fu,Su),d(Gxn,Txn)).
Letting n→∞, we have
(34)d(Su,Fu)≤ϕ(d(Fu,Fu),d(Fu,Su),d(Fu,Fu))=ϕ(0,d(Fu,Su),0)<d(Fu,Su),impliesthat,Fu=Su.
The weak compatibility ofF and S implies that FSu=SFu and then SSu=SFu=FSu=FFu.
On the other hand, sinceS(X)⊆G(X), there exists a v∈X such that Su=Gv.
Now, we show thatGv=Tv.
From (5), we have
(35)d(Su,Tv)≤ϕ(d(Fu,Gv),d(Fu,Su),d(Gv,Tv))=ϕ(0,0,d(Gv,Tv))<d(Gv,Tv),impliesthatGv=Tv,thatis,Su=Fu=Gv=Tv.
The weak compatibility of T and G implies that TGv=GTv and GGv=GTv=TGv=TTv.
Let us show thatSu is the common fixed point of S,  T, F and G.
From (5), we have
(36)d(SSu,Tv)≤ϕ(d(FSu,Gv),d(FSu,SSu),d(Gv,Tv))=ϕ(d(SSu,Tv),0,0)<d(SSu,Tv),impliesthatSSu=Tv=Su,thatis,Su=SSu=FSu.
Therefore, Su is the common fixed point of S and F.
Similarly, one can prove thatTv is the common fixed point of T and G.
SinceSu=Tv, we conclude that Su is the common fixed point of S,  T, F, and G.
The proof is similar whenG(X) is assumed to be complete subspace of X. The cases in which S(X) or T(X) is a complete subspace of X are similar to the cases in which G(X) or F(X), respectively, is complete since S(X)⊆G(X) and T(X)⊆F(X).
IfSu=Tu=Fu=Gu=u and Sv=Tv=Fv=Gv=v, then (5) gives
(37)d(Su,Tv)≤ϕ(d(Fu,Gv),d(Fu,Su),d(Gv,Tv))<d(Fu,Gv),impliesthatu=v.
Therefore,u=v and the common fixed point is unique.


## 4. (CLR) Property

Theorem 14.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+→ℝ+ be a 𝒲  function. Let S,  T,  F, and  G be self-maps on X satisfying (3), (6), and the following:
(38)T(X)⊆F(X)andthepair{T,G}satisfies(CLRG)property,
or
(39)S(X)⊆G(X)andthepair{S,F}satisfies(CLRF)property.
Then S, T, F, and G have a unique common fixed point.Proof.
Without loss of generality, we assume thatT(X)⊆F(X) and the pair {T,G} satisfies (CLRG) property; then there exists some sequence {xn} in X such that Txn and Gxn converge to Gx, for some x in X as n→∞.
SinceT(X)⊆F(X) there exists a sequence {yn} in X such that Txn=Fyn; hence Fyn→Gx as n→∞.
Now, we claim thatlimn→∞Syn=Gx.
Letlimn→∞Syn=z.
Takingx=yn and y=xn in (3), we have
(40)d(Syn,Txn)≤ϕ(max{d(Fyn,Gxn),d(Fyn,Syn),d(Gxn,Txn)}).
Letting n→∞, we have
(41)d(z,Gx)≤ϕ(max{d(Gx,Gx),d(Gx,z),d(Gx,Gx)})=ϕ(d(Gx,z))<d(z,Gx),impliesthatz=Gx.
Subsequently, we have Txn, Gxn, Fyn, and Syn that converge to z.
Now, we claim thatTx=z.
Takingx=yn and y=x in (3), we have
(42)d(Syn,Tx)≤ϕ(max{d(Fyn,Gx),d(Fyn,Syn),d(Gx,Tx)}).
Letting n→∞, we have
(43)d(z,Tx)≤ϕ(max{d(Gx,Gx),d(z,z),d(z,Tx)})=ϕ(d(z,Tx))<d(z,Tx),impliesthatz=Tx=Gx.
Since the pair (T,G) is weakly compatible, it follows that Tz=Gz.
Also, sinceT(X)⊆F(X), there exists some y in X such that Tx=Fy(=z).
We next show thatFy=Sy(=z).
Takingy=xn, x=y in (3), we have
(44)d(Sy,Txn)≤ϕ(max{d(Fy,Gxn),d(Fy,Sy),d(Gxn,Txn)}).
Letting n→∞, we have
(45)d(Sy,z)≤ϕ(max{d(z,z),d(z,Sy),d(z,z)})=ϕ(d(z,Sy))<d(Sy,z),impliesthatSy=z=Fy.
But the pair {S,F} is weakly compatible; it follows that Sz=Fz.
Next, we claim thatSz=Tz.
Takingx=z,  y=z in (3), we have
(46)d(Sz,Tz)≤ϕ(max{d(Fz,Gz),d(Fz,Sz),d(Gz,Tz)})=ϕ(d(Sz,Tz))<d(Sz,Tz),impliesthatSz=Tz,thatis,Sz=Tz=Fz=Gz.
Now, we show thatz=Sz.
Takingx=z,  y=x in (3), we have
(47)d(Sz,Tx)≤ϕ(max{d(Fz,Gx),d(Fz,Sz),d(Gx,Tx)})=ϕ(d(Sz,z))<d(Sz,z),impliesthatz=Sz=Tz=Fz=Gz.
Hence z is the common fixed point of S, T, F, and G.
IfSz=Tz=Fz=Gz=z and St=Tt=Ft=Gt=t, then (3) gives
(48)d(Sz,Tt)≤ϕ(max{d(Fz,Gt),d(Fz,Sz),d(Gt,Tt)})=ϕ(d(Sz,Tt))<d(Sz,Tt),impliesthatSz=Tt,thatis,z=t.
Therefore, uniqueness follows.Theorem 15.
Let(X,d) be Hausdorff and complete g.m.s. and let ϕ:ℝ+3→ℝ+ be a 𝒮 function. Let S, T, F, and G be self-maps on X satisfying (5), (6), and (38). Then S, T, F, and G have a unique common fixed point.Proof.
Without loss of generality, we assume thatT(X)⊆F(X) and the pair {T,G} satisfies (CLRG) property; then there exists some sequence {xn} in X such that Txn and Gxn converge to Gx, for some x in X as n→∞.
SinceT(X)⊆F(X) there exists a sequence {yn} in X such that Txn=Fyn; hence Fyn→Gx as n→∞.
Now, we claim thatlimn→∞Syn=Gx.
Letlimn→∞Syn=z.
Takingx=yn and y=xn in (5), we have
(49)d(Syn,Txn)≤ϕ(d(Fyn,Gxn),d(Fyn,Syn),d(Gxn,Txn)).
Letting  n→∞, we have
(50)d(z,Gx)≤ϕ(d(Gx,Gx),d(Gx,z),d(Gx,Gx))=ϕ(0,d(Gx,z),0)<d(z,Gx),impliesthatz=Gx.
Subsequently, we have Txn, Gxn, Fyn, and Syn that converge to z.
Now, we claim thatTx=z.
Takingx=yn and y=x in (5), we have
(51)d(Syn,Tx)≤ϕ(d(Fyn,Gx),d(Fyn,Syn),d(Gx,Tx)).
Letting n→∞, we have
(52)d(z,Tx)≤ϕ(d(Gx,Gx),d(z,z),d(z,Tx))=ϕ(0,0,d(z,Tx))<d(z,Tx),impliesthatz=Tx=Gx.
Since the pair (T,G) is weakly compatible, it follows that Tz=Gz.
Also, sinceT(X)⊆F(X), there exists some y in X such that Tx=Fy(=z).
We next show thatFy=Sy(=z).
Takingy=xn, x=y in (5), we have
(53)d(Sy,Txn)≤ϕ(d(Fy,Gxn),d(Fy,Sy),d(Gxn,Txn)).
Letting n→∞, we have
(54)d(Sy,z)≤ϕ(d(z,z),d(z,Sy),d(z,z))=ϕ(0,d(z,Sy),0)<d(Sy,z),impliesthatSy=z=Fy.
But the pair {S,F} is weakly compatible; it follows that Sz=Fz.
Next, we claim thatSz=Tz.
Takingx=z, y=z in (5), we have
(55)d(Sz,Tz)≤ϕ(d(Fz,Gz),d(Fz,Sz),d(Gz,Tz))=ϕ(d(Sz,Tz),0,0)<d(Sz,Tz),impliesthatSz=Tz.
Hence, Sz=Tz=Fz=Gz.
Now, we show thatz=Sz.
Takingx=z, y=x in (5), we have
(56)d(Sz,Tx)≤ϕ(d(Fz,Gx),d(Fz,Sz),d(Gx,Tx))=ϕ(d(Sz,z),0,0)<d(Sz,z),impliesthatz=Sz=Tz=Fz=Gz.
Hence z is the common fixed point of S,   T,  F and G.
IfSz=Tz=Fz=Gz=z and St=Tt=Ft=Gt=t, then (5) gives
(57)d(Sz,Tt)≤ϕ(d(Fz,Gt),d(Fz,Sz),d(Gt,Tt))=ϕ(d(Sz,Tt),0,0)<d(Sz,Tt),impliesthatSz=Ttorz=t.
Therefore, z=t and the common fixed point is unique.



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*Source: 719324-2013-04-16.xml*