# Common Fixed Point Theorems for a Rational Inequality in Complex Valued Metric Spaces

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

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## Abstract

We prove a common fixed point theorem for a pair of mappings. Also, we prove a common fixed point theorem for pairs of self-mappings along with weakly commuting property.


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## Body

## 1. Introduction

Azam et al. [1] introduced the notion of complex valued metric spaces and established some fixed point theorems for the mappings satisfying a rational inequality. The definition of a cone metric space banks on the underlying Banach space which is not a division Ring. The idea of rational expressions is not meaningful in cone metric spaces, and therefore many results of analysis cannot be generalized to cone metric spaces. The complex valued metric spaces form a special class of cone metric space, and we can study improvements of host results of analysis involving divisions.A complex numberz∈ℂ is an ordered pair of real numbers, whose first coordinate is called Re(z) and second coordinate is called Im(z).Letℂ be the set of complex numbers and z1, z2∈ℂ. Define a partial order ≾ on ℂ as follows: z1≾z2 if and only if Re(z1)≤Re(z2) and Im(z1)≤Im(z2); that is, z1≾z2, if one of the following holds:(C1)
Re(z1)=Re(z2) and Im(z1)=Im(z2);(C2)
Re(z1)<Re(z2) and Im(z1)=Im(z2);(C3)
Re(z1)=Re(z2) and Im(z1)<Im(z2);(C4)
Re(z1)<Re(z2) and Im(z1)<Im(z2).In particular, we will write z1⋨z2 if z1≠z2 and one of (C2), (C3), and (C4) is satisfied, and we will write z1≺z2 if only (C4) is satisfied.Remark 1.
We note that the following statements hold:(i)
a, b∈ℝ and a≤b⇒az≾bz  ∀z∈ℂ,(ii)
0≾z1⋨z2⇒|z1|<|z2|,(iii)
z1≾z2 and z2≺z3⇒z1≺z3.Azam et al. [1] defined the complex valued metric space (X, d) as follows.Definition 2.
LetX be a nonempty set. Suppose that the mapping d:X×X→ℂ satisfies the following conditions:(i)
0≾d(x,y), for all x, y∈X and d(x,y)=0 if and only if x=y;(ii)
d(x,y)=d(y,x)   for all  x, y∈X;(iii)
d(x,y)≾d(x,z)+d(z,y),  for all  x, y, z∈X.Then, d is called a complex valued metric on X, and (X,d) is called a complex valued metric space.Example 3.
LetX=ℂ. Define the mapping d:X×X→ℂ by
(1)d(x,y)=𝒾|x-y|,∀x,y∈X.
Then, (X,d) is a complex valued metric space.Definition 4.
Let(X,d) be a complex valued metric space. A sequence {xn} in X is said to be (i)
convergent tox, if for every c∈ℂ with 0≺c there is k∈ℕ such that, for all n>k, d(xn,x)≺c. We denote this by {xn}→x as n→∞ or limn→∞xn=x;(ii)
Cauchy, if for everyc∈ℂ with 0≺c there is k∈ℕ such that for all n>k, d(xn,xn+m)≺c, where m∈ℕ;(iii)
complete, if every Cauchy sequence inX converges in X.Lemma 5.
Let(X,d) be a complex valued metric space, and let {xn} be a sequence in X. Then, {xn} converges to x if and only if |d(xn,x)|→0 as n→∞.Lemma 6.
Let(X,d) be a complex valued metric space, and let {xn} be a sequence in X. Then, {xn} is a Cauchy sequence if and only if |d(xn,xn+m)|→0 as n→∞, where m∈ℕ.In 1982, Sessa [2] introduced the notion of weak commutativity as follows.Definition 7.
Two self-mapsf and g of a metric space (X,d) are said to be weakly commuting if d(fgx,gfx)≤d(fx,gx), for all x in X.In a similar mode, we introduce the notion of weak commutativity in complex valued metric spaces as follows.Definition 8.
Two self mapsf and g of a complex valued metric space (X,d) are said to be weakly commuting if d(fgx,gfx)≾d(fx,gx),  for all x in X.Example 9.
LetX=[0,∞), and define d:X×X→ℂ by d(x,y)=𝒾|x-y|, for all x, y∈X.
Then,(X,d) is a complex valued metric space.
Definefx=x and gx=2x.
Then, clearlyd(fgx,gfx)≾d(fx,gx), for all xinX.
Thus,f and g are weakly commuting.


## 2. Main Theorem

Theorem 10.
Letf and g be self mappings of a complex valued metric space (X,d) satisfying the following:
(2)d(fx,gy)≾k[d(x,fx)d(x,gy)+{d(x,y)}2+d(x,fx)d(x,y)d(x,fx)+d(x,y)+d(x,gy)]∀x,yinXwithx≠y,0<k<1,d(x,fx)+d(x,y)+d(x,gy)≠0.
Then, f and g have a common fixed point.
Further, ifd(x,fx)+d(x,y)+d(x,gy)=0 implies that d(fx,gy)=0, then f and g have a unique common fixed point.Proof.
Letx0∈X. Define a sequence {xn} in X by x2n+1=fx2n, x2n+2=gx2n+1, n=0,1,2,….
Letd(x,fx)+d(x,y)+d(x,gy)≠0.
From (2), we have
(3)d(x2n+1,x2n+2)=d(fx2n,gx2n+1)≾k[(d(x2n,fx2n)+d(x2n,x2n+1)+d(x2n,gx2n+1))-1(d(x2n,fx2n)d(x2n,gx2n+1)+{d(x2n,x2n+1)}2+d(x2n,fx2n)d(x2n,x2n+1)d(x2n,fx2n)d(x2n,gx2n+1)+{d(x2n,x2n+1)}2)×(d(x2n,fx2n)+d(x2n,x2n+1)+d(x2n,gx2n+1))-1]=k[×(d(x2n,x2n+1)+d(x2n,x2n+1)+d(x2n,x2n+2))-1(d(x2n,x2n+1)d(x2n,x2n+2)+{d(x2n,x2n+1)}2+d(x2n,x2n+1)d(x2n,x2n+1)d(x2n,fx2n)d(x2n,gx2n+1)+{d(x2n,x2n+1)}2)×(d(x2n,x2n+1)+d(x2n,x2n+1)+d(x2n,x2n+2))-1]=kd(x2n,x2n+1)×[d(x2n,x2n+2)+d(x2n,x2n+1)+d(x2n,x2n+1)d(x2n,x2n+1)+d(x2n,x2n+1)+d(x2n,x2n+2)]=kd(x2n,x2n+1).
Similarly, we have(4)d(x2n,x2n+1)=d(fx2n,gx2n-1)≾k[×(d(x2n,fx2n)+d(x2n,x2n-1)+d(x2n,gx2n-1))-1(d(x2n,fx2n)d(x2n,gx2n-1)+{d(x2n,x2n-1)}2+d(x2n,fx2n)d(x2n,x2n-1)d(x2n,fx2n)d(x2n,gx2n+1)+{d(x2n,x2n+1)}2)×(d(x2n,fx2n)+d(x2n,x2n-1)+d(x2n,gx2n-1))-1]=k[×(d(x2n,x2n+1)+d(x2n,x2n-1)+d(x2n,x2n))-1(d(x2n,x2n+1)d(x2n,x2n)+{d(x2n,x2n-1)}2+d(x2n,x2n+1)d(x2n,x2n-1)d(x2n,fx2n)d(x2n,gx2n+1)+{d(x2n,x2n+1)}2)×(d(x2n,x2n+1)+d(x2n,x2n-1)+d(x2n,x2n))-1]=kd(x2n,x2n-1)[d(x2n,x2n-1)+d(x2n,x2n+1)d(x2n,x2n+1)+d(x2n,x2n-1)]=kd(x2n,x2n-1)=kd(x2n-1,x2n).
Consequently, it can be concluded that
(5)d(xn,xn+1)≾kd(xn-1,xn)≾k2d(xn-2,xn-1)⋮≾knd(x0,x1).
Now, for all m>n,
(6)d(xm,xn)≾d(xn,xn+1)+d(xn+1,xn+2)+⋯+d(xm,xm-1)≾knd(x0,x1)+kn+1d(x0,x1)+⋯+km-1d(x0,x1)≾kn1-kd(x0,x1).
Therefore, we have
(7)|d(xm,xn)|≤kn1-k|d(x0,x1)|.
Hence,
(8)limn→∞|d(xm,xn)|=0.
Hence,{xn} is a Cauchy sequence in X. But X is complete metric space, so {xn} is convergent to some point, say z, in X, that is, xn→z as n→∞.
Now, we will prove thatz=gz.
Let, if possible,z≠gz.
Now, using the triangular inequality and (2), we have
(9)d(z,gz)≾d(z,x2n+1)+d(x2n+1,gz)=d(z,x2n+1)+d(fx2n,gz)≾d(z,x2n+1)+k[×(d(x2n,fx2n)+d(x2n,z)+d(x2n,gz))-1(d(x2n,fx2n)d(x2n,gz)+{d(x2n,z)}2+d(x2n,fx2n)d(x2n,z)+{d(x2n,x2n+1)}2)×(d(x2n,fx2n)+d(x2n,z)+d(x2n,gz))-1]=d(z,x2n+1)+k[×(d(x2n,x2n+1)+d(x2n,z)+d(x2n,gz))-1(d(x2n,x2n+1)d(x2n,gz)+{d(x2n,z)}2+d(x2n,x2n+1)d(x2n,z)+{d(x2n,x2n+1)}2)×(d(x2n,x2n+1)+d(x2n,z)+d(x2n,gz))-1].
Thus, we have
(10)|d(z,gz)|≤|d(z,x2n+1)|+k|[×(d(x2n,x2n+1)+d(x2n,z)+d(x2n,gz))-1(d(x2n,x2n+1)d(x2n,gz)+{d(x2n,z)}2+d(x2n,x2n+1)d(x2n,z)+{d(x2n,x2n+1)}2)×(d(x2n,x2n+1)+d(x2n,z)+d(x2n,gz))-1]|.
Lettingn→∞, we have |d(z,gz)|≤0, a contradiction.
Hence, we getz=gz; that is, z is the fixed point of g.
Again assume thatz≠fz.
From (2) and using the triangular inequality, we have
(11)d(z,fz)≾d(z,x2n+2)+d(x2n+2,fz)=d(z,x2n+2)+d(fz,x2n+2)=d(z,x2n+2)+d(fz,gx2n+1)≾d(z,x2n+2)+k[×(d(z,fz)+d(z,x2n+1)+d(z,gx2n+1))-1d(z,gx2n+1)-1(d(z,fz)d(z,gx2n+1)+{d(z,x2n+1)}2+d(z,fz)d(z,x2n+1)+{d(x2n,x2n+1)}2)×(d(z,fz)+d(z,x2n+1)+d(z,gx2n+1))-1]=d(z,x2n+2)+k[×(d(z,fz)+d(z,x2n+1)+d(z,x2n+2))-1(d(z,fz)d(z,x2n+2)+{d(z,x2n+1)}2+d(z,fz)d(z,x2n+1)+{d(x2n,x2n+1)}2)×(d(z,fz)+d(z,x2n+1)+d(z,x2n+2))-1].
Thus, we have
(12)|d(z,gz)|≤|d(z,x2n+2)|+k|[×(d(z,fz)+d(z,x2n+1)+d(z,x2n+2))-1(+d(z,fz)d(z,x2n+1)d(z,fz)d(z,x2n+2)+{d(z,x2n+1)}2+d(z,fz)d(z,x2n+1)+{d(x2n,x2n+1)}2)×(d(z,fz)+d(z,x2n+1)+d(z,x2n+2))-1]|.
Lettingn→∞, we have |d(z,gz)|≤0, a contradiction.
Hence, we getz=fz; that is, z is a fixed point of f.
Therefore, we find thatz is a common fixed point of f and g.Uniqueness.Let w(≠z) be another fixed point of g. Suppose that d(x,fx)+d(x,y)+d(x,gy)=0 implies d(fx,gy)=0.Now,(13)d(z,fz)+d(z,w)+d(z,gw)=0impliesd(fz,gw)=0.
Therefore, we get
(14)d(z,w)=d(fz,gw)=0,whichimpliesthatz=w.
Hence, f and g have a unique common fixed point.Corollary 11.
Letf be a self-map of a complex valued metric space (X,d) satisfying the following:
(15)d(fx,fy)≾k[d(x,fx)d(x,fy)+{d(x,y)}2+d(x,fx)d(x,y)d(x,fx)+d(x,y)+d(x,fy)],∀x,yinXwithx≠y,0<k<1,d(x,fx)+d(x,y)+d(x,fy)≠0.
Then, f has a fixed point.
Further, ifd(x,fx)+d(x,y)+d(x,fy)=0 implies that d(fx,fy)=0 then f has a unique common fixed point.Proof.
By puttingf=g in Theorem 10, we get Corollary 11.


## 3. Weakly Commuting Property

Theorem 12.
LetA, B, S, and T be self mappings of a complex valued metric space (X,d) satisfying the following:(3.1)
SX⊆ BX, TX ⊆ AX,(3.2)
the pairs(A,S) and (B,T) are weakly commuting,(3.3)
for allx, y in X, either
(16)d(Sx,Ty)≾α[×(d(Ax,Sx)+d(Ax,By)+d(Ax,Ty))-1(d(Ax,Sx)d(Ax,Ty)+{d(Ax,By)}2+d(Ax,Sx)d(Ax,By)+{d(Ax,By)}2)×(d(Ax,Sx)+d(Ax,By)+d(Ax,Ty))-1]+βd(Ax,By),if d(Ax,Sx)+d(Ax,By)+d(Ax,Ty)≠0, where |α+β| < 1 and |β|<1;  d(Sx,Ty)=0, if d(Ax,Sx)+d(Ax,By)+d(Ax,Ty)=0.
If any ofA, B, S, or T is continuous, then A, B, S, and T have a unique common fixed point z.Proof.
Letx0∈X. Since SX⊆BX, so there exists a point x1 in X such that Sx0=Bx1. Also, since TX⊆AX, we can choose a point x2 in X such that Tx1=Ax2.
Continuing this process, we haveSx2n=Bx2n+1 and Tx2n+1=Ax2n+2, for n=0,1,2,….
Defined2n=d(Sx2n,Tx2n+1) and d2n+1=d(Sx2n+2,Tx2n+1).
Supposed2n≠0 and d2n+1≠0 for =1,2,3,….
From (3.3), we have(17)d2n+1=d(Sx2n+2,Tx2n+1)≾α[+d(Ax2n+2,Tx2n+1))-1(+{d(Ax2n+2,Bx2n+1)}2d(Ax2n+2,Sx2n+2)d(Ax2n+2,Tx2n+1)+{d(Ax2n+2,Bx2n+1)}2+d(Ax2n+2,Sx2n+2)d(Ax2n+2,Bx2n+1)+{d(Ax2n+2,Bx2n+1)}2)×(+d(Ax2n+2,Tx2n+1))d(Ax2n+2,Sx2n+2)+d(Ax2n+2,Bx2n+1)+d(Ax2n+2,Tx2n+1))-1]+βd(Ax2n+2,Bx2n+1)=α[+d(Tx2n+1,Sx2n)+d(Tx2n+1,Tx2n+1))-1(+{d(Tx2n+1,Sx2n)}2d(Tx2n+1,Sx2n+2)d(Tx2n+1,Tx2n+1)+{d(Tx2n+1,Sx2n)}2+d(Tx2n+1,Sx2n+2)d(Tx2n+1,Sx2n)+{d(Tx2n+1,Sx2n)}2)×(+d(Tx2n+1,Sx2n)+d(Tx2n+1,Tx2n+1))d(Tx2n+1,Sx2n+2)+d(Tx2n+1,Sx2n)+d(Tx2n+1,Tx2n+1))-1]+βd(Tx2n+1,Sx2n)=α[d(Tx2n+1,Sx2n)-1(d(Tx2n+1,Sx2n)×{d(Tx2n+1,Sx2n)+d(Tx2n+1,Sx2n+2)})×(d(Tx2n+1,Sx2n+2)+d(Tx2n+1,Sx2n))-1]+βd(Tx2n+1,Sx2n)=(α+β)d(Tx2n+1,Sx2n)=(α+β)d2n.
Thus, we have
(18)|d2n+1|≤|α+β||d2n|=k|d2n|,wherek=|α+β|<1.
In general, we have(19)|d2n+1|≤k|d2n|≤k2|d2n-1|⋮≤k2n+1|d0|,thatis,|d2n+1|≤k2n+1|d0|.
Letting n→∞, we have
(20)|d2n+1|≤0,whichimpliesthatd2n+1=0.
Therefore, d(Sx2n+2,Tx2n+1)→0 as n→∞.
We get the following sequence:(3.4)
{Sx0,Tx1,Sx2,Tx3,…,Sx2n,Tx2n+1,…}, which is a Cauchy sequence in the complete complex valued metric space (X,d), therefore converges to a limit point z in X.
Therefore, the sequences{Sx2n}={Bx2n+1} and {Tx2n-1}={Ax2n}, which are the subsequences of (3.4) hence also converge to the same point z in X.
Now, suppose thatA is continuous so that the sequences {A2x2n} and {ASx2n} converge to the same point Az. Since A and S are weakly commuting, we have
(21)d(SAx2n,ASx2n)≾d(Ax2n,Sx2n),whichimpliesthat|d(SAx2n,ASx2n)|≤|d(Ax2n,Sx2n)|.
Lettingn→∞, we have
(22)limn→∞|d(SAx2n,Az)|≤|d(z,z)|=0,whichimpliesthatlimn→∞|d(SAx2n,Az)|=0;thatis,SAx2n⟶zasn⟶∞.
Now, we will show thatAz=z. Let, if possible, Az≠z.
Now, using the triangle inequality and (3.3), we get(23)d(Az,z)≾d(Az,SAx2n)+d(SAx2n,Tx2n+1)+d(Tx2n+1,z)≾d(Az,SAx2n)+α[+d(A2x2n,Tx2n+1))-1(+{d(A2x2n,Bx2n+1)}2d(A2x2n,SAx2n)d(A2x2n,Tx2n+1)+{d(A2x2n,Bx2n+1)}2+d(A2x2n,SAx2n)d(A2x2n,Bx2n+1)+{d(A2x2n,Bx2n+1)}2)×(+d(A2x2n,Tx2n+1))d(A2x2n,SAx2n)+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))-1]+βd(A2x2n,Bx2n+1)+d(Tx2n+1,z).
Thus, we have
(24)|d(Az,z)|≤|d(Az,SAx2n)|+|α[(+{d(A2x2n,Bx2n+1)}2d(A2x2n,SAx2n)d(A2x2n,Tx2n+1)+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))-1+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))-1+{d(A2x2n,Bx2n+1)}2+d(A2x2n,SAx2n)d(A2x2n,Bx2n+1)+{d(A2x2n,Bx2n+1)}2)×(+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))d(A2x2n,SAx2n)+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))-1]+βd(A2x2n,Bx2n+1)+d(A2x2n,Bx2n+1)+d(A2x2n,Tx2n+1))-1|+|d(Tx2n+1,z)|.
Lettingn→∞, we have
(25)|d(Az,z)|≤|d(Az,Az)|+|α[×(d(Az,Az)+d(Az,z)+d(Az,z))-1(d(Az,Az)d(Az,z)+{d(Az,z)}2+d(Az,Az)d(Az,z)+{d(Az,z)}2)×(d(Az,Az)+d(Az,z)+d(Az,z))-1]+βd(Az,z)×(d(Az,Az)+d(Az,z)+d(Az,z))-1|+|d(z,z)|=|α2+β||d(Az,z)|,whichimpliesthat|α2+β|≥1,acontradictionto|α+β|<1.
Hence, Az=z.
Now, we will prove thatSz=z.
Again, using the triangle inequality and (3.3), we have(26)d(Sz,z)≾d(Sz,Tx2n+1)+d(Tx2n+1,z)≾α[(d(Az,Sz)+d(Az,Bx2n+1)+d(Az,Tx2n+1))-1(d(Az,Sz)d(Az,Tx2n+1)+{d(Az,Bx2n+1)}2+d(Az,Sz)d(Az,Bx2n+1){d(Az,Bx2n+1)}2)×(+d(Az,Tx2n+1))d(Az,Sz)+d(Az,Bx2n+1)+d(Az,Tx2n+1))-1]+βd(Az,Bx2n+1)+d(Tx2n+1,z).
Thus, we have
(27)|d(Sz,z)|≤|α[+d(Az,Bx2n+1)+d(Az,Tx2n+1))-1(d(Az,Sz)d(Az,Tx2n+1)+{d(Az,Bx2n+1)}2+d(Az,Sz)d(Az,Bx2n+1)+{d(Az,Bx2n+1)}2)×(+d(Az,Bx2n+1)+d(Az,Tx2n+1))d(Az,Sz)+d(Az,Bx2n+1)+d(Az,Tx2n+1))-1]+βd(Az,Bx2n+1)+d(Az,Bx2n+1)+d(Az,Tx2n+1))-1|+|d(Tx2n+1,z)|.
Letting n→∞, we have
(28)|d(Sz,z)|≤|α[d(z,Sz)d(z,z)+{d(z,z)}2+d(z,Sz)d(z,z)d(z,Sz)+d(Az,z)+d(z,z)]+βd(z,z)d(z,Sz)d(z,z)+{d(z,z)}2+d(z,Sz)d(z,z)d(z,Sz)+d(Az,z)+d(z,z)|+|d(z,z)|=0,whichimpliesthatSz=z.
Now, since SX⊆BX, there exists a point w in X such that Bw=z.
Thus, we have(29)d(z,Tw)=d(Sz,Tw)≾α[×(d(Az,Sz)+d(Az,Bw)+d(Az,Tw))-1(d(Az,Sz)d(Az,Tw)+{d(Az,Bw)}2+d(Az,Sz)d(Az,Bw)+{d(Az,Bw)}2)×(d(Az,Sz)+d(Az,Bw)+d(Az,Tw))-1]+βd(Az,Bw),whichimpliesthat|d(z,Tw)|≤α·0+β·0=0,thatis,Tw=z.
SinceB and T are weakly commuting, so we have
(30)d(TBw,BTw)≾d(Bw,Tw),thatis,|d(TBw,BTw)|≤|d(Bw,Tw)|,whichimpliesthat|d(Tz,Bz)|≤|d(z,z)|=0;thatis,Tz=Bz.
Now, we will prove that Tz=z. Let, if possible, Tz≠z.
From (3.3), we have(31)d(z,Tz)=d(Sz,Tz)≾α[(d(Az,Sz)+d(Az,Bz)+d(Az,Tz))-1(d(Az,Sz)d(Az,Tz)+{d(Az,Bz)}2+d(Az,Sz)d(Az,Bz){d(Az,Bz)}2)×(d(Az,Sz)+d(Az,Bz)+d(Az,Tz))-1]+βd(Az,Bz).
Thus, we have(32)|d(z,Tz)|≤|α2+β||d(z,Tz)|,whichimpliesthat|α2+β|≥1,acontradictionto|α+β|<1.
Hence, Tz=z=Bz and Sz=z=Az.
So,z is the common fixed point of A, B, S, and T.
Now, if one of the mappingsB, S, or T is continuous instead of A, then one can show that A, B, S, and T have a common fixed point.
To show thatz is unique, let u be another common fixed point of A and S.
From (3.3), we have(33)d(u,z)=d(Su,Tz)≾α[×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1(d(Au,Su)d(Au,Tz)+{d(Au,Bz)}2+d(Au,Su)d(Au,Bz){d(Au,Bz)}2)×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1]+βd(Au,Bz);thatis,|d(u,z)|≤|×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1α[×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1(d(Au,Su)d(Au,Tz)+{d(Au,Bz)}2+d(Au,Su)d(Au,Bz)+{d(Au,Bz)}2)×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1]+βd(Au,Bz)×(d(Au,Su)+d(Au,Bz)+d(Au,Tz))-1|,whichimpliesthat|d(u,z)|≤|α2+β||d(u,z)|,impliesthat,|α2+β|≥1,acontradictionto|α+β|<1.
Thus, we have, u=z that is, A and S have a unique common fixed point.
In the same way, it can be shown thatz is the unique common fixed point of B and T.



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*Source: 942058-2013-10-24.xml*