# Some Common Fixed Point Theorems in Partial Metric Spaces

Authors: Erdal Karapınar; Uğur Yüksel
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

Many problems in pure and applied mathematics reduce to a problem
of common fixed point of some self-mapping operators which are defined
on metric spaces. One of the generalizations of metric spaces is the partial
metric space in which self-distance of points need not to be zero but the
property of symmetric and modified version of triangle inequality is satisfied. In this paper, some well-known results on common fixed point are
investigated and generalized to the class of partial metric spaces.


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## Body

## 1. Introduction and Preliminaries

Partial metric spaces, introduced by Matthews [1, 2], are a generalization of the notion of the metric space in which in definition of metric the condition d(x,x)=0 is replaced by the condition d(x,x)≤d(x,y). Different approaches in this area have been reported including applications of mathematical techniques to computer science [3–7].In [2], Matthews discussed some properties of convergence of sequences and proved the fixed point theorems for contractive mapping on partial metric spaces: any mapping T of a complete partial metric space X into itself that satisfies, where 0≤k<1, the inequality d(Tx,Ty)≤kd(x,y), for all x,y∈X, has a unique fixed point. Recently, many authors (see e.g., [8–16]) have focused on this subject and generalized some fixed point theorems from the class of metric spaces to the class of partial metric spaces.The definition of partial metric space is given by Matthews (see e.g., [1]) as follows.Definition 1.1.
LetX be a nonempty set and let p:X×X→ℝ0+ satisfy
(1.1)(PM1)x=y⟺p(x,x)=p(y,y)=p(x,y),(PM2)p(x,x)≤p(x,y),(PM3)p(x,y)=p(y,x),(PM4)p(x,y)≤p(x,z)+p(z,y)-p(z,z)
for all x, y, and z∈X, where ℝ0+=[0,∞). Then the pair (X,p) is called a partial metric space (in short PMS) and p is called a partial metric on X.Let(X,p) be a PMS. Then, the functions dp,dm:X×X→ℝ0+ given by(1.2)dp(x,y)=2p(x,y)-p(x,x)-p(y,y),dm(x,y)=max⁡⁡{p(x,y)-p(x,x),p(x,y)-p(y,y)}
are (usual) metrics on X. It is clear that dp and dm are equivalent. Each partial metric p on X generates a T0 topology τp on X with a base of the family of open p-balls {Bp(x,ɛ):x∈X,ɛ>0}, where Bp(x,ɛ)={y∈X:p(x,y)<p(x,x)+ɛ} for all x∈X and ɛ>0. A basic example of partial metric is (ℝ0+,p), where p(x,y)=max{x,y}. One can easily deduce that dp(x,y)=|x-y|=dm(x,y).Example 1.2 (See [1, 2]).
LetX={[a,b]:a,b∈ℝ,a≤b} and define p([a,b],[c,d])=max{b,d}-min{a,c}. Then (X,p) is a partial metric spaces.We give same topological definitions on partial metric spaces.Definition 1.3 (see e.g., [1, 2, 13]).
(i) A sequence{xn} in a PMS (X,p) converges to x∈X if and only if p(x,x)=limn→∞p(x,xn).
(ii) A sequence{xn} in a PMS (X,p) is called a Cauchy sequence if and only if limn,m→∞p(xn,xm) exists (and finite).
(iii) A PMS(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=limn,m→∞p(xn,xm).
(iv) A mappingf:X→X is said to be continuous at x0∈X if for every ɛ>0, there exists δ>0 such that f(B(x0,δ))⊂B(f(x0),ɛ).The following lemmas will be frequently used in the proofs of the main results.Lemma 1.4 (see e.g., [1, 2, 13]).
(A) A sequence{xn} is Cauchy in a PMS (X,p) if and only if {xn} is Cauchy in a metric space (X,dp).
(B) A PMS(X,p) is complete if and only if the metric space (X,dp) is complete. Moreover,
(1.3)lim⁡n→∞dp(x,xn)=0⟺p(x,x)=lim⁡n→∞p(x,xn)=lim⁡n,m→∞p(xn,xm),
where x is a limit of {xn} in (X,dp).Remark 1.5.
Let(X,p) be a PMS. Therefore,  (A)
ifp(x,y)=0, then x=y;(B)
ifx≠y, then p(x,y)>0.
This follows immediately from the definition and can easily be verified by the reader.Lemma 1.6 (See e.g., [15]).
Assumexn→z as n→∞ in a PMS (X,p) such that p(z,z)=0. Then limn→∞p(xn,y)=p(z,y) for every y∈X.In this paper, we extend some common fixed point theorems for two self-mappings without commuting property from the class of usual metric spaces (see e.g., [17]) to the class of partial metric spaces (see Theorems 2.6 and 2.2). We also consider some common fixed point theorems with commuting property on partial metric spaces (see Theorem 2.7, Corollary 2.8).


## 2. Main Results

We first recall the definition of a common fixed point of two self-mappings.Definition 2.1.
Let(X,p) be a PMS and S,T two self-mappings on (X,p). A point z∈X is said to be a common fixed point of S and T if Sz=Tz=z.In the sequel, we give the first results about a common fixed point theorem. We prove the existence and uniqueness of a common fixed point of two self-mappings under certain conditions. Notice that here the operators need not commute with each other.Theorem 2.2.
Suppose that(X,p) is a complete PMS and T, S are self-mappings on X. If there exists an r∈[0,1) such that
(2.1)p(Tx,Sy)≤rM(x,y)
for any x,y∈X, where
(2.2)M(x,y)=max⁡⁡{p(Tx,x),p(Sy,y),p(x,y),12[p(Tx,y)+p(Sy,x)]},
then there exists z∈X such that Tz=Sz=z.Proof.
Letx0∈X. Define the sequence {xn}n=1∞ in a way that x2=Tx1 and x1=Sx0 and inductively
(2.3)x2k+2=Tx2k+1,x2k+1=Sx2kfork=0,1,2,….
If there exists a positive integer N such that x2N=x2N+1, then x2N is a fixed point of S and hence a fixed point of T. Indeed, since x2N=x2N+1=Sx2N, then
(2.4)Sx2N=Sx2N+1=S2x2N⟹x2N=Sx2N=Sx2N+1=x2N+1.
Also, due to (2.1) we have
(2.5)p(x2N+2,x2N+1)=p(Tx2N+1,Sx2N)≤rM(x2N+1,x2N),
where
(2.6)M(x2N+1,x2N)=max⁡⁡{12p(Tx2N+1,x2N+1),p(Sx2N,x2N),p(x2N+1,x2N),12[p(Tx2N+1,x2N)+p(Sx2N,x2N+1)]}=max⁡⁡{12p(Tx2N+1,x2N+1),p(x2N,x2N),p(x2N+1,x2N),12[p(Tx2N+1,x2N)+p(x2N+1,x2N+1)]}=p(Tx2N+1,x2N+1)=p(Tx2N+1,x2N)=p(x2N+2,x2N+1).
Thus we have (2.5) implies (1-r)p(x2N+2,x2N+1)≤0. Since r<1, then p(x2N+2,x2N+1)=0, which yields that Tx2N+1=x2N+2=x2N+1. Notice that x2N+1=x2N is the fixed point of S. As a result, x2N+1=x2N is the common fixed point of S and T. A similar conclusion holds if x2N+1=x2N+2 for some positive integer N. Therefore, we may assume that xk≠xk+1 for all k. If k is odd, due to (2.1), we have
(2.7)p(xk+1,xk+2)=p(Txk,Sxk+1)≤rM(xk,xk+1),
where
(2.8)M(xk,xk+1)=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1),p(xk,xk+1),12[p(xk+1,xk+1)+p(xk+2,xk)]}.
In view of (PM4), we have
(2.9)p(xk+1,xk+1)+p(xk+2,xk)≤p(xk+2,xk+1)+p(xk+1,xk).
Thus, (2.8) turns into
(2.10)M(xk,xk+1)=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1),p(xk,xk+1),12[p(xk+2,xk+1)+p(xk+1,xk)]}=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1)}.
IfM(xk,xk+1)=p(xk+2,xk+1), then since r<1, the inequality (2.7) yields a contradiction. Hence, M(xk,xk+1)=p(xk+1,xk) and by (2.7) we have
(2.11)p(xk+2,xk+1)≤rp(xk+1,xk).
Ifk is even, the same inequality (2.11) can be obtained analogously.
We get that{p(xk,xk+1)} is a nonnegative, nonincreasing sequence of real numbers. Regarding (2.11), one can observe that
(2.12)p(xk,xk+1)≤rkp(x0,x1),∀k=0,1,2,….
Consider now(2.13)dp(xk+1,xk+2)=2p(xk+1,xk+2)-p(xk+1,xk+1)-p(xk+2,xk+2)≤2p(xk+1,xk+2)≤2rk+1p(x0,x1).
Hence, regarding (2.12), we have limk→∞dp(xk+1,xk+2)=0. Moreover,
(2.14)dp(xk+1,xk+s)≤dp(xk+s-1,xk+s)+⋯+dp(xk+1,xk+2)≤2rk+sp(x0,x1)+⋯+2rn+1p(x0,x1).
After standard calculation, we obtain that {xk} is a Cauchy sequence in (X,dp) that is, dp(xk,xm)→0 as k,m→∞. Since (X,p) is complete, by Lemma 1.4, (X,dp) is complete and the sequence {xk} is convergent in (X,dp) to, say, z∈X.
Again by Lemma1.4,
(2.15)p(z,z)=lim⁡k→∞p(xk,z)=lim⁡k,m→∞p(xk,xm).
Since {xk} is a Cauchy sequence in (X,dp), we have limk,m→∞dp(xk,xm)=0. We assert that limk,m→∞p(xk,xm)=0. Without loss of generality, we assume that n>m. Now observe that
(2.16)p(xn+2,xn)≤p(xn+2,xn+1)+p(xn+1,xn)-p(xn+1,xn+1)≤p(xn+2,xn+1)+p(xn+1,xn).
Analogously,
(2.17)p(xn+3,xn)≤p(xn+3,xn+2)+p(xn+2,xn)-p(xn+2,xn+2)≤p(xn+3,xn+2)+p(xn+2,xn).
Taking into account (2.16), the expression (2.17) yields
(2.18)p(xn+3,xn)≤p(xn+3,xn+2)+p(xn+2,xn+1)+p(xn+1,xn).
Inductively, we obtain
(2.19)p(xm,xn)≤p(xm,xm-1)+⋯+p(xn+2,xn+1)+p(xn+1,xn).
Due to (2.12), the expression (2.19) turns into
(2.20)p(xm,xn)≤rm-1p(x1,x0)+⋯+rn+1p(x1,x0)+rnp(x1,x0)≤rn(1+r+rm-n-1)p(x1,x0).
Regarding r<1, by simple calculations, one can observe that
(2.21)lim⁡k,m→∞p(xk,xm)=0.
Therefore, from (2.15), we have
(2.22)p(z,z)=lim⁡k→∞p(xk,z)=lim⁡k,m→∞p(xk,xm)=0.
We assert thatTz=z. On the contrary, assume Tz≠z. Then p(z,Tz)>0. Let {x2k(i)} be a subsequence of {x2k} and hence of {xk}. Due to (2.1), we have
(2.23)p(Sx2k(i),Tz)≤rM(x2k(i),z),
where
(2.24)M(x2k(i),z)=max⁡⁡{p(x2k(i),x2k(i)+1),p(Tz,z),p(x2k(i),z),12[p(Tz,x2k(i))+p(z,x2k(i)+1)]}.
Letting k→∞ and taking into account (2.22), the expression (2.24) implies that
(2.25)M(x2k(i),z)=max⁡⁡{0,p(Tz,z),0,12p(Tz,z)}=p(Tz,z).
Thus,
(2.26)p(z,Tz)≤rp(Tz,z).
Since r<1, we have p(Tz,z)=0. By Remark 1.5, we get Tz=z. Analogously, if we choose a subsequence {x2k(i)+1} of {x2k+1}, we obtain Sz=z. Hence Tz=Sz=z.Remark 2.3.
We notice that Theorem2.2 can be obtained from Theorem 2.1 in [18] or Theorem 5 in [19] by simple manipulations.
However, explicit proof of Theorem2.2 has a crucial role in the proofs of Proposition 2.5, Theorems 2.6 and 2.7.
These results cannot be obtained from the mentioned papers [18, 19].Example 2.4.
LetX=[0,1] and p(x,y)=max{x,y}. Then (X,p) is a complete PMS. Clearly, p is not a metric. Suppose S,T:X→X such that Sx=Tx=x/3 and r=1/2. Without loss of generality, assume x≥y. Then
(2.27)p(Tx,Sy)=max⁡⁡{x3,y3}=x3≤12M(x,y),
where
(2.28)M(x,y)=max⁡⁡{x,y,x,12[x+p(y,Sx)]}=x.
Thus, all conditions of Theorem 2.2 are satisfied, and 0 is the common fixed point of S and T.Proposition 2.5.
Suppose(X,p) is a complete PMS and T, S are self-mappings on X. If there exists an r∈[0,1) such that
(2.29)p(Tmx,Sny)≤rM(x,y)
for any x,y∈X and some positive integers m,n, where
(2.30)M(x,y)=max⁡⁡{p(Tmx,x),p(Sny,y),p(x,y),12[p(Tmx,y)+p(Sny,x)]},
then there exists z∈X such that Tmz=Snz=z.Proof.
Letx0∈X. As in the proof of the previous theorem, we define a sequence {xn}n=1∞ in a way that x2=Tmx1 and x1=Snx0; we get inductively
(2.31)x2k+2=Tmx2k+1,x2k+1=Snx2kfork=0,1,2,….
If there exists a positive integer N such that x2N=x2N+1, then x2N is a fixed point of Tm and hence a fixed point of Sn. A similar conclusion holds if x2N+1=x2N+2 for some positive integer N. Therefore, we may assume that xk≠xk+1 for all k.
Ifk is odd, due to (2.29), we have
(2.32)p(xk+1,xk+2)=p(Tmxk,Snxk+1)≤rM(xk,xk+1),
where
(2.33)M(xk,xk+1)=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1),p(xk,xk+1),12[p(xk+1,xk+1)+p(xk+2,xk)]}.
By means of (PM4), we have
(2.34)p(xk+1,xk+1)+p(xk+2,xk)≤p(xk+2,xk+1)+p(xk+1,xk).
Thus, (2.33) becomes
(2.35)M(xk,xk+1)=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1),p(xk,xk+1),12[p(xk+2,xk+1)+p(xk+1,xk)]}=max⁡⁡{p(xk+1,xk),p(xk+2,xk+1)}.
IfM(xk,xk+1)=p(xk+2,xk+1), then since r<1, the inequality (2.32) yields a contradiction. Hence, M(xk,xk+1)=p(xk+1,xk) and by (2.32), we have
(2.36)p(xk+2,xk+1)≤rp(xk+1,xk).
Ifk is even, analogously we obtain the same inequality (2.36).
We obtain that{p(xk,xk+1)} is a nonnegative, nonincreasing sequence of real numbers. Regarding (2.36), one has
(2.37)p(xk,xk+1)≤rkp(x0,x1),∀k=0,1,2,….
Consider now(2.38)dp(xk+1,xk+2)=2p(xk+1,xk+2)-p(xk+1,xk+1)-p(xk+2,xk+2)≤2p(xk+1,xk+2)≤2rk+1p(x0,x1).
Hence, regarding (2.37), we have limk→∞dp(xk+1,xk+2)=0. Moreover,
(2.39)dp(xk+1,xk+s)≤dp(xk+s-1,xk+s)+⋯+dp(xk+1,xk+2)≤2rk+sp(x0,x1)+⋯+2rk+1p(x0,x1)
which implies that {xk} is a Cauchy sequence in (X,dp), that is, dp(xk,xm)→0 as k,m→∞. Since (X,p) is complete, by Lemma 1.4, (X,dp) is complete and the sequence {xk} is convergent in (X,dp) to, say, z∈X.
By Lemma1.4,
(2.40)p(z,z)=lim⁡k→∞p(xk,z)=lim⁡k,m→∞p(xk,xm).
Since {xk} is a Cauchy sequence in (X,dp), we have
(2.41)lim⁡k,m→∞dp(xk,xm)=0.
We claim that limk,m→∞p(xk,xm)=0. Following the steps (2.16)–(2.22) in the proof of Theorem 2.2, we conclude the result. Thus,
(2.42)2p(xm,xn)-p(xm,xm+1)-p(xn,xn+1)≤2p(xm,xn)-p(xm,xm)-p(xn,xn)=dp(xm,xn).
Thus, lettingn,m→∞ in view of (2.37), (2.41), the expression (2.42) yields that limk,m→∞p(xk,xm)=0. Therefore, from (2.40) we have
(2.43)p(z,z)=lim⁡k→∞p(xk,z)=lim⁡k,m→∞p(xk,xm)=0.
We assert thatTmz=z. Assume the contrary, that is, Tmz≠z, then p(z,Tmz)>0. Let {x2k(i)} be a subsequence of {x2k} and hence of {xk}. Due to (2.29), we have
(2.44)p(Snx2k(i),Tmz)≤rM(x2k(i),z)≤rmax⁡⁡{12p(x2k(i),x2k(i)+1),p(Tmz,z),p(x2k(i),z),12[p(Tmz,x2k(i))+p(z,x2k(i)+1)]}.
Lettingk→∞ and taking into account (2.43), the expression (2.44) implies that
(2.45)p(z,Tmz)≤rmax⁡⁡{0,p(Tmz,z),0,12p(Tmz,z)}≤rp(Tmz,z).
Since r<1, we have p(Tmz,z)=0. By Remark 1.5, we get Tmz=z. Analogously, if we choose a subsequence {x2k(i)+1} of {x2k+1}, we obtain Snz=z. Hence Tmz=Snz=z.The following theorem is a generalization of a common fixed point theorem that requires no commuting criteria (see e.g., [17]).Theorem 2.6.
Suppose(X,p) is a complete PMS and T,S are self-mappings on X. If there exists an r∈[0,1) such that
(2.46)p(Tmx,Sny)≤rM(x,y)
for any x,y∈X and some positive integers m,n, where
(2.47)M(x,y)=max⁡⁡{p(Tmx,x),p(Sny,y),p(x,y),12[p(Tmx,y)+p(Sny,x)]},
then T and S have a unique common fixed point z∈X.Proof.
Due to Proposition2.5, we have
(2.48)Tmz=Snz=z.
We claim thatz is a common fixed point of S and T. From (2.46) and (2.48), it follows that
(2.49)p(Tz,z)=p(TTmz,Snz)=p(TmTz,Snz)≤rM(Tz,z),
where
(2.50)M(Tz,z)=max⁡⁡{p(TmTz,Tz),p(Snz,z),p(Tz,z),12[p(TmTz,z)+p(Snz,Tz)]}=max⁡⁡{p(TTmz,Tz),p(z,z),p(Tz,z),12[p(TTmz,z)+p(z,Tz)]}=max⁡⁡{p(Tz,Tz),p(Tz,z),12[p(Tz,z)+p(z,Tz)]}=max⁡⁡{p(Tz,Tz),p(Tz,z)}.
Due to (PM3), we have p(Tz,Tz)≤p(Tz,z). Hence
(2.51)M(Tz,z)=max⁡⁡{p(Tz,Tz),p(Tz,z)}=p(Tz,z).
Regarding the assumption r<1 and the expression (2.49), we get p(Tz,z)≤rp(Tz,z) which implies that p(Tz,z)=0, and by Remark 1.5, we obtain Tz=z. Analogously, one can show that Sz=z. Hence, Tz=Sz=z.
For the uniqueness of the common fixed pointz, assume the contrary. Suppose w is another common fixed point of S and T. Then,
(2.52)p(z,w)=p(Tmz,Snw)≤rM(z,w),
where
(2.53)M(z,w)=max⁡⁡{p(Tmz,z),p(Snw,w),p(z,w),12[p(Tmz,w)+p(Snw,z)]}=max⁡⁡{p(z,z),p(w,w),p(z,w),12[p(z,w)+p(w,z)]}=p(z,w).
Therefore, p(z,w)≤rp(z,w). Since 0≤r<1, one has p(z,w)=0 which yields z=w by Remark 1.5. Hence, z is a unique common fixed point of S and T.Theorem 2.7.
Let(X,p) be a complete PMS. Suppose that T,S,F, and G are self-mappings on X, and F and G are continuous. Suppose also that T,F and S,G are commuting pairs and that
(2.54)T(X)⊂F(X),S(X)⊂G(X).
If there exists an r∈[0,1), and m,n∈ℕ such that
(2.55)p(Tx,Sy)≤rM(x,y)
for any x, y in X, where
(2.56)M(x,y)=max⁡⁡{p(Tx,Gx),p(Sy,Fy),p(Gx,Fy),12[p(Tx,Fy)+p(Sy,Gx)]},
then T, S, F, and G have a unique common fixed point z in X.Proof.
Fixx0∈X. Since T(X)⊂F(X) and S(X)⊂G(X), we can choose x1, x2 in X such that y1=Fx1=Tx0 and y2=Gx2=Sx1. In general, we can choose x2n-1, x2n in X such that
(2.57)y2n-1=Fx2n-1=Tx2n-2,y2n=Gx2n=Sx2n-1,n=1,2,….
We claim that the constructive sequence{yn} is a Cauchy sequence. If there exists a positive integer N such that yN=yN+1, then yN=yN+1=yN+2=⋯=yN+k for all k∈ℕ. Therefore, {yn} is a Cauchy sequence and we proved claim. Thus, we may assume that yn≠yn+1 for all n.
By (2.55) and (2.57),
(2.58)p(y2n+1,y2n+2)=p(Fx2n+1,Gx2n+2)=p(Tx2n,Sx2n+1)≤rM(x2n,x2n+1),
where
(2.59)M(x2n,x2n+1)=max⁡⁡{12p(Tx2n,Gx2n),p(Sx2n+1,Fx2n+1),p(Gx2n,Fx2n+1),12[p(Tx2n,Fx2n+1)+p(Sx2n+1,Gx2n)]}=max⁡⁡{12p(Tx2n,Sx2n-1),p(Sx2n+1,Tx2n),p(Sx2n-1,Tx2n),12[p(Tx2n,Tx2n)+p(Sx2n+1,Sx2n-1)]}.
Due to (PM4), we have(2.60)p(Sx2n-1,Sx2n+1)+p(Tx2n,Tx2n)≤p(Sx2n-1,Tx2n)+p(Sx2n+1,Tx2n).
Hence
(2.61)M(x2n,x2n+1)=max⁡⁡{p(Tx2n,Sx2n-1),p(Sx2n+1,Tx2n)}.
But if M(x2n,x2n+1)=p(Sx2n+1,Tx2n), then by (2.58)
(2.62)p(Sx2n+1,Tx2n)≤rp(Sx2n+1,Tx2n),r∈[0,1),
which implies p(Sx2n+1,Tx2n)=0. Thus, M(x2n,x2n+1)=p(Sx2n-1,Tx2n), and consequently,
(2.63)p(Sx2n+1,Tx2n)≤rp(Sx2n-1,Tx2n),
or, equivalently,
(2.64)p(y2n+2,y2n+1)≤rp(y2n+1,y2n).
Analogously, one can show that
(2.65)p(y2n+3,y2n+2)≤rp(y2n+2,y2n+1).
Indeed, from (2.55) and (2.57),
(2.66)p(y2n+3,y2n+2)=p(Sx2n+1,Tx2n+2)≤rM(x2n+1,x2n+2),
where
(2.67)M(x2n+2,x2n+1)=max⁡⁡{12p(Tx2n+1,Gx2n+1),p(Sx2n+2,Fx2n+2),p(Gx2n+1,Fx2n+2),12[p(Tx2n+1,Fx2n+2)+p(Sx2n+2,Gx2n+1)]}=max⁡⁡{12p(Tx2n+2,Sx2n+1),p(Sx2n+1,Tx2n),p(Sx2n+1,Tx2n),12[p(Tx2n+2,Tx2n)+p(Sx2n+1,Sx2n+1)]}=max⁡⁡{p(Tx2n+2,Sx2n+1),p(Sx2n+1,Tx2n)}.
If M(x2n+2,x2n+1)=p(Tx2n+2,Sx2n+1)=p(y2n+3,y2n+2), then by (2.66), we have a contradiction. Thus M(x2n+2,x2n+1)=p(Sx2n+1,Tx2n)=p(y2n+2,y2n+1) which proves (2.65).
Thus, we conclude thatp(yn+1,yn)≤rp(yn,yn-1), for all n∈ℕ.
By elementary calculation, regarding0<r<1, we conclude that {yn} is a Cauchy sequence. Since {X,d} is complete, {yn} converges to a point z∈X. Consequently, the subsequences {Tmx2n}, {Snx2n-1}, {Gx2n}, and {Fx2n-1} converge to z.
Regarding thatT, F and S, G are commuting pairs and the continuity of G and F, the sequences {FFx2n-1},{SFx2n-1} tend to Fz, and the sequences {GGx2n},{TGx2n} tend to Gz, as n→∞.
Thus,(2.68)p(Gz,Fz)=lim⁡n→∞p(TGx2n,SFx2n-1)≤rlim⁡n→∞M(Gx2n,Fx2n-1),
where
(2.69)M(Gx2n,Fx2n-1)=max⁡⁡{12p(TGx2n,GGx2n),p(SFx2n-1,FFx2n-1),p(GGx2n,FFx2n-1),12[p(TGx2n,FFx2n-1)+p(GGx2n,SnFx2n-1)]}.
Sincelimn→∞M(Gx2n,Fx2n-1)=p(Gz,Fz), one has p(Gz,Fz)≤rp(Gz,Fz), that is, Gz=Fz. Analogously, one obtains
(2.70)Tz=Sz=Fz=Gz.
To conclude the proof, consider
(2.71)p(z,Fz)=lim⁡n→∞p(Fx2n-1,SFx2n-1)=lim⁡n→∞p(Tx2n-2,SFx2n-1)≤rlim⁡n→∞M(x2n-2,Fx2n-1),
where
(2.72)M(x2n-2,Fx2n-1)=max⁡⁡{12p(Tx2n-2,Gx2n-2),p(SFx2n-1,FFx2n-1),p(Gx2n-2,FFx2n-1),12[p(Tx2n-2,FFx2n-1)+p(SFx2n-1,Gx2n-2)]},
Letting n→∞ in (2.72) and having in mind (2.70), we get that limn→∞M(x2n-2,Fx2n-1)=p(z,Fz). Due to (2.71), we have Fz=z. Thus, we have
(2.73)Tz=Sz=Fz=Gz=z.
We assert thatz is unique. Suppose on the contrary that there is another common fixed point w of S, T, F, and G. Then p(z,w)=p(Tz,Sw)≤rM(z,w), where
(2.74)M(z,w)=max⁡⁡{p(Tz,Gz),p(Sw,Fw),p(Gz,Fw),12[p(Tz,Fw)+p(Sw,Gz)]}=max⁡⁡{p(z,z),p(w,w),p(z,w),12[p(z,w)+p(w,z)]}=p(z,w).
Since M(z,w)=p(z,w),
(2.75)p(z,w)≤rp(z,w).
Therefore, p(z,w)=0, and by Remark 1.5, we have z=w. Hence z is the unique common fixed point of S, T, F, and G.Regarding the relation between Theorems2.2 and 2.6, one concludes the following corollary in view of Theorem 2.7.Corollary 2.8.
Let(X,p) be a complete PMS. Suppose that A,B,F, and G are self-mappings on X, and F and G are continuous. Suppose also that A,B, and F,G are commuting pairs and also A and G commutes each other and
(2.76)A(X)⊂F(X),B(X)⊂G(X).
If there exists r∈[0,1), and m,n∈ℕ such that
(2.77)p(Amx,Bny)≤rM(x,y)
for any x, y in X, where
(2.78)M(x,y)=max⁡⁡{p(Amx,Gx),p(Bny,Fy),p(Gx,Fy),12[p(Amx,Fy)+p(Bny,Gx)]},
then A, B, F, and G have a unique common fixed point z in X.Proof.
Due to Theorem2.7,
(2.79)Amz=Bnz=Fz=Gz=z.
Following the steps of the proof of Theorem 2.7 with T=Am, S=Bn, we get (2.70) which is equivalent to (2.79). Thus, Am, Bn, F, and G have a unique common fixed point z in X.
We claim that(2.80)Az=Bz=z.
By (2.79),
(2.81)p(Az,z)=p(AAmz,Bnz)=p(AmAz,Bnz)≤rlim⁡n→∞M(Az,z),
where
(2.82)M(Az,z)=max⁡⁡{p(AmAz,GAz),p(Bnz,Fz),p(GAz,Fz),12[p(AmAz,Fz)+p(Bnz,GAz)]}=max⁡⁡{p(AAmz,AGz),p(z,z),p(AGz,z),12[p(AAmz,z)+p(z,AGz)]}=max⁡⁡{p(Az,Az),0,p(Az,z),12[p(Az,z)+p(z,Az)]}=p(Az,z).
Hence, (2.81) is equivalent to p(Az,z)≤rp(Az,z) which yields p(Az,z)=0, that is, Az=z. Analogously, one can get Bz=z.  Indeed, By (2.79),
(2.83)p(z,Bz)=p(Amz,BBnz)=p(Amz,BnBz)≤rlim⁡n→∞M(z,Bz),
where
(2.84)M(z,Bz)=max⁡⁡{p(Amz,Gz),p(BnBz,FBz),p(Gz,FBz),12[p(Amz,FBz)+p(BnBz,Gz)]}=max⁡⁡{p(z,z),p(Bz,FBz),p(z,FBz),12[p(z,FBz)+p(Bz,z)]}=max⁡⁡{0,p(Bz,Bz),p(z,Bz),12[p(Bz,z)+p(z,Bz)]}=p(Bz,z).
Hence, (2.83) is equivalent to p(Bz,z)≤rp(Bz,z) which yields p(z,Bz)=0, that is, z=Bz.  Hence,
(2.85)Az=Bz=z.
Combining (2.80) and (2.85), we obtain Gz=Fz=Az=Bz=z.



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*Source: 263621-2011-11-28.xml*