Proof.
Let x0 be an arbitrary point in X. Since f(X)⊆T(X), we can find x1∈X such that fx0=Tx1 and also, as gx1∈S(X), there exist x2∈X such that gx1=Sx2. In general, x2n+1∈X is chosen such that fx2n=Tx2n+1 and x2n+2∈X such that gx2n+1=Sx2n+2, we obtain a sequences {yn} in X such that
(2.3)y2n=fx2n=Tx2n+1, y2n+1=gx2n+1=Sx2n+2, ∀n≥0.
Suppose y2m=y2m+1 for some m. Thus, g and T have a coincidence point. Due to (2.1), we have
(2.4)p(y2m+2,y2m+1)=p(fx2m+2,gx2m+1)≤δM(x2m+2,x2m+1)+LN(x2m+2,x2m+1),
where
(2.5)N(x2m+2,x2m+1)=min {pw(fx2m+2,Sx2m+2),pw(gx2m+1,Tx2m+1),pw(Sx2m+2,gx2m+1),pw(fx2m+2,Tx2m+1)}=min {pw(y2m+2,y2m+1),pw(y2m+1,y2m),pw(y2m+1,y2m+1),pw(y2m+2,y2m)}=0,M(x2m+2,x2m+1)=max {p(Sx2m+2,Tx2m+1),p(fx2m+2,Sx2m+2),p(gx2m+1,Tx2m+1),p(Sx2m+2,gx2m+1)+p(fx2m+2,Tx2m+1)2}=max {p(y2m+1,y2m),p(y2m+2,y2m+1),p(y2m+1,y2m),p(y2m+1,y2m+1)+p(y2m+2,y2m)2}=p(y2m+2,y2m+1).
So,
(2.6)p(y2m+2,y2m+1)≤δp(y2m+2,y2m+1).
Therefore, by δ∈[0,1[, we have p(y2m+2,y2m+1)=0, that is, y2m+1=y2m+2. So, f and S have a coincidence point.
Suppose now that yn≠yn+1 for all n≥0. From (2.1), we obtain
(2.7)p(y2n,y2n+1)=p(fx2n,gx2n+1)≤δM(x2n,x2n+1)+LN(x2n,x2n+1),
where
(2.8)N(x2n,x2n+1)=min {pw(fx2n,Sx2n),pw(gx2n+1,Tx2n+1),pw(Sx2n,gx2n+1),pw(fx2n,Tx2n+1)}=min {pw(y2n,y2n-1),pw(y2n+1,y2n),pw(y2n-1,y2n+1),pw(y2n,y2n)}=0,M(x2n,x2n+1)=max {p(Sx2n,Tx2n+1),p(fx2n,Sx2n),p(gx2n+1,Tx2n+1),p(Sx2n,gx2n+1)+p(fx2n,Tx2n+1)2}=max {p(y2n-1,y2n),p(y2n,y2n-1),p(y2n+1,y2n),p(y2n-1,y2n+1)+p(y2n,y2n)2}.
Due to (2.7), we have
(2.9)p(y2n,y2n+1)≤δM(x2n,x2n+1).
Due to PM4, we have
(2.10)p(y2n-1,y2n+1)+p(y2n,y2n)≤p(y2n-1,y2n)+p(y2n,y2n+1).
Hence, M(x2n,x2n+1)=max {p(y2n,y2n-1),p(y2n+1,y2n)}. If M(x2n,x2n+1)=p(y2n+1,y2n), then by (2.7)
(2.11)p(y2n+1,y2n)≤δp(y2n+1,y2n).
Since δ∈[0,1[, the inequality (2.9) yields a contradiction. Hence, M(x2n,x2n+1)=p(y2n,y2n-1), then by (2.7) we have
(2.12)p(y2n+1,y2n)≤δp(y2n,y2n-1).
Thus, one can observe that
(2.13)p(yn+1,yn)≤δnp(y0,y1), ∀n=0,1,2,….
Consider now
(2.14)ps(yn+2,yn+1)=2p(yn+2,yn+1)-p(yn+2,yn+2)-p(yn+1,yn+1)≤2p(yn+2,yn+1)≤δn+1p(y0,y1).
Hence, regarding (2.13), we have
(2.15)lim n→∞ps(yn+2,yn+1)=0.
Moreover,
(2.16)ps(yn+1,yn+k)≤ps(yn+k-1,yn+k)+⋯+ps(yn+1,yn+2)≤2δn+k-1p(y0,y1)+⋯+2δn+1p(y0,y1).
After standard calculation, we obtain that {yn} is a Cauchy sequence in (X,ps), that is, ps(yn,ym)→0 as n, m→∞. Since (X,p) is complete, by Lemma 1.4, (X,ps) is complete and sequence {yn} is convergent in (X,ps) to say z∈X. From Lemma 1.4,
(2.17)p(z,z)=lim n→∞p(yn,z)=lim n,m→∞p(yn,ym).
Since {yn} is a Cauchy sequence in (X,ps), we have
(2.18)lim n,m→∞ps(yn,ym)=0.
We assert that lim n,m→∞ p(yn,ym)=0. Without loss of generality, we assume that n>m,
(2.19)p(yn+2,yn)≤p(yn+2,yn+1)+p(yn+1,yn)-p(yn+1,yn+1)≤p(yn+2,yn+1)+p(yn+1,yn).
Similarly,
(2.20)p(yn+3,yn)≤p(yn+3,yn+2)+p(yn+2,yn)-p(yn+2,yn+2)≤p(yn+3,yn+2)+p(yn+2,yn).
Taking into account (2.20), the expression (2.19) yields
(2.21)p(yn+3,yn)≤p(yn+3,yn+2)+p(yn+2,yn+1)+p(yn+1,yn).
Inductively, we obtain
(2.22)p(ym,yn)≤p(ym,ym+1)+⋯+p(yn-2,yn-1)+p(yn-1,yn).
Due to (2.13),
(2.23)p(ym,yn)≤δmp(y0,y1)+⋯+δn-2p(y0,y1)+δn-1p(y0,y1)≤δm(1+δ+⋯+δn-m-1)p(y0,y1).
Regarding δ∈[0,1[, we can observe that lim n,m→∞p(yn,ym)=0.
Since yn→z in X, {fx2n}, {Tx2n+1}, {gx2n+1}, {Sx2n+2} converge to z.
Now we show that z is the fixed point for maps g and T. Assume that T(X) is complete, there exists u∈X such that z=Tu. We will show that gu=z. On the contrary, assume that gu≠z.
From, (2.1) we have
(2.24)p(fx2n,gu)≤δM(x2n,u)+LN(x2n,u),
where
(2.25)N(x2n,u)=min {pw(fx2n,Sx2n),pw(gu,Tu),pw(Sx2n,gu),pw(fx2n,Tu)}=min {pw(fx2n,Sx2n),pw(gu,z),pw(Sx2n,gu),pw(fx2n,z)},M(x2n,u)=max {p(Sx2n,Tu),p(fx2n,Sx2n),p(gu,Tu),p(Sx2n,gu)+p(fx2n,Tu)2}=max {p(Sx2n,z),p(fx2n,Sx2n),p(gu,z),p(Sx2n,gu)+p(fx2n,z)2}.
Since lim n→∞M(x2n,u)=p(gu,z) and lim n→∞N(x2n,u)=0. We get
(2.26)p(z,gu)≤δp(gu,z).
Since δ∈[0,1[, we get p(z,gu)=0. Therefore, gu=Tu=z. Since the maps g and T are weakly compatible, we have gz=gTu=Tgu=Tz. We will also show that gz=z. From (2.1), we have
(2.27)p(fx2n,gz)≤δM(x2n,z)+LN(x2n,z),
where
(2.28)N(x2n,z)=min {pw(fx2n,Sx2n),pw(gz,Tz),pw(Sx2n,gz),pw(fx2n,Tz)},M(x2n,z)=max {p(Sx2n,Tz),p(fx2n,Sx2n),p(gz,Tz),p(Sx2n,gz)+p(fx2n,Tz)2}=max {p(Sx2n,gz),p(fx2n,Sx2n),p(gz,gz),p(Sx2n,gz)+p(fx2n,Tz)2}.
Since lim n→∞M(x2n,z)=p(z,gz) and lim n→∞N(x2n,z)=0, then
(2.29)p(z,gz)=lim n→∞p(fx2n,gz)≤δp(z,gz).
Since δ∈[0,1[, p(z,gz)=0. By Remark 1.5, we get z=gz.
Similarly, we show that z is also fixed point of f and S. Hence, fz=gz=Tz=Sz=z.
The proofs for the cases in which S(X), f(X), or g(X) is complete are similar.
Last, we show z is unique. Suppose on the contrary that there is another common fixed point t of f, g, S, and T. Then
(2.30)p(z,t)=p(fz,gt)≤δM(z,t)+LN(z,t),
where
(2.31)N(z,t)=min {pw(fz,Sz),pw(gt,Tt),pw(Sz,gt),pw(fz,Tt)}=0,M(z,t)=max {p(Sz,Tt),p(fz,Sz),p(gt,Tt),p(Sz,gt)+p(fz,Tt)2}=p(Sz,Tt)=p(z,t).
Thus,
(2.32)p(z,t)≤δp(z,t).
Therefore, p(z,t)=0 and Remark 1.5 z=t. So, z is the unique common fixed point os f, g, S, and T.