Proof.
Suppose M(x,y)=0 for some x,y∈X.
Then fx∈Sx and gy∈Ty. Thus x is a coincidence point of f and S, and y is a coincidence point of g and T.
Assume that M(x,y)>0 for all x,y∈X.
Let ϵ=(1/2)(1/(s2+s)-q) and β=q+ϵ=(1/2)(1/(s2+s)+q).
Then ϵ>0 and 0<β<1.
Let x0∈X. From (2.1.2), there exist x1,y1∈X such that y1=gx1∈Sx0.
From (2.1.2), there exist x2,y2∈X such that y2=fx2∈Tx1 and from Lemma 5, we have
(13)d(y1,y2)≤H(Sx0,Tx1)+ϵM(x0,x1)≤qM(x0,x1)+ϵM(x0,x1), from (2.1.1)=βM(x0,x1).
Again from (2.1.2), there exist x3,y3∈X such that y3=gx3∈Sx2 and from Lemma 5, we have
(14)d(y2,y3)≤H(Sx2,Tx1)+ϵM(x2,x1)≤qM(x2,x1)+ϵM(x2,x1), from (2.1.1)=βM(x2,x1).
Continuing in this way, we get the sequences {xn} and {yn} in X such that y2n+1=gx2n+1∈Sx2n, y2n+2=fx2n+2∈Tx2n+1 for all n=0,1,2,3,… and
(15)d(y2n+1,y2n)≤βM(x2n,x2n-1),d(y2n+1,y2n+2)≤βM(x2n,x2n+1) for n=1,2,3,….
For simplification, write dn=d(yn,yn+1).
Now
(16)d2n+1=d(y2n+1,y2n+2)≤βM(x2n,x2n+1)=βmax{d(y2n,y2n+1),d(y2n,y2n+1), d(y2n+1,y2n+2),d(y2n,y2n+2),0}≤βmax{d2n,d2n+1,s[d2n+d2n+1]}
which in turn yields
(17)d2n+1≤γd2n,
where γ=max{β,sβ/(1-sβ)}.
Similarly, we can show that
(18)d2n≤γd2n-1.
Thus, from (17) and (18), we have dn≤γdn-1 for all n=1,2,3,…, so that
(19)dn≤γnd0 ∀n=1,2,3,….
If γ=β, then γs=βs=(s/2)(1/(s2+s)+q)<s/(s2+s)=1/(s+1)≤1/2<1.
If γ=sβ/(1-sβ), then γs=s(sβ/(1-sβ))<s[1/(s+1)/1-1/(s+1)]=1.
Thus
(20)γs<1.
Since γ<1, from (19), we have
(21)dn⟶0 as n⟶∞.
Now for m,n∈ℕ with m>n, we have
(22)d(yn,ym)≤sd(yn,yn+1) +s2d(yn+1,yn+2)+⋯+sm-nd(ym-1,ym) ≤(sγn+s2γn+1+⋯+sm-nγm-1)d0, from (19) ≤sγn(1+sγ+⋯+sm-n-1γm-n-1+⋯)d0 =γns1-γsd0→0 as n,m→∞.
Hence {yn} is a Cauchy sequence in X. Since X is complete, there exists z∈X such that yn→z which in turn yields fx2n→z and gx2n+1→z as n→∞.
Suppose the mappings (f,g;S) satisfy the condition (W.C.C). Then
(23)d(fx,gy)≤d(y,Sx) ∀x,y∈X.
From (23), we have
(24)d(fx2n,gz)≤d(z,Sx2n)≤d(z,gx2n+1).
Letting n→∞, we get
(25)d(z,gz)≤0 so that gz=z.
Now
(26)1sd(z,Tz) ≤d(z,gx2n+1)+d(gx2n+1,Tz) ≤d(z,gx2n+1)+H(Sx2n,Tz) ≤d(z,y2n+1) +qmax{d(fx2n,gz),d(fx2n,Sx2n),d(gz,Tz),d(fx2n,Tz),d(gz,Sx2n)} ≤d(z,y2n+1) +qmax{d(y2n,z),d(y2n,y2n+1),d(z,Tz),d(y2n,Tz),d(z,y2n+1)}.
Using (vii) of Lemma 4 and letting n→∞ and using (21), we get
(27)1sd(z,Tz)≤0+qsd(z,Tz).
Thus we have
(28)d(z,Tz)≤qs2d(z,Tz),
so that
(29)d(z,Tz)=0 since qs2<1.
From Lemma 6, we have z∈Tz. Thus
(30)gz=z∈Tz.
From (23), we get
(31)d(fz,z)=d(fz,gz)≤d(z,Sz).
Now from (2.1.1), we have
(32)d(z,Sz)≤H(Sz,Tz)≤qmax{d(fz,gz),d(fz,Sz),d(gz,Tz), d(fz,Tz),d(gz,Sz)}≤qmax{d(z,Sz),s[d(z,Sz)+d(z,Sz)], 0,sd(z,Sz),d(z,Sz)} from (30) and (31)=2qsd(z,Sz).
Hence z∈Sz, since 2qs<1.
From (31), we have fz=z. Thus
(33)fz=z∈Sz.
From (30) and (33), it follows that z is a common fixed point of f, g, S, and T.